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All questions of Binomial Theorem for Mathematics Exam
The fourth and fifth term of a sequence {tn}n≥1 are 4 and 5 respectively and the nth term is given as tn = 2tn-1 - tn-2, n > 3 (n ∈ N). Then the sum to first 2009 terms is - a)2019045
- b)2013021
- c)2017036
- d)20180 40
Correct answer is option 'A'. Can you explain this answer?
The fourth and fifth term of a sequence {tn}n≥1 are 4 and 5 respectively and the nth term is given as tn = 2tn-1 - tn-2, n > 3 (n ∈ N). Then the sum to first 2009 terms is
a)
2019045
b)
2013021
c)
2017036
d)
20180 40
|
Chirag Verma answered |
Thus {an} is a constant sequence
a5 = t5 − t4 = 1
Now a4 = t4 − t3 ⇒ 1 = 4 − t3 ⇒ t3 = 3
Similarly t2 = 2 , t1 = 1
Thus {tn} is an A.P with r = 1 and common difference 1
23C0 + 23C2+ 23C4 + .. + 23C22 equals - a)223 – 2
- b)222
- c)211
- d)
Correct answer is option 'B'. Can you explain this answer?
23C0 + 23C2+ 23C4 + .. + 23C22 equals
a)
223 – 2
b)
222
c)
211
d)
|
Veda Institute answered |
Given sum = sum of odd terms 
The middle term in the expansion of (1 – 2x + x2)n is - a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
The middle term in the expansion of (1 – 2x + x2)n is
a)
b)
c)
d)
|
|
Chirag Verma answered |
Here 2n is even integer, therefore,
term will be the middle term.
The middle term in the expansion of (1 – 2x + x2)n is - a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
The middle term in the expansion of (1 – 2x + x2)n is
a)
b)
c)
d)
|
|
Chirag Verma answered |
Here 2n is even integer, therefore,
term will be the middle term.
The number of rational terms in the expansion of 
- a)6
- b)7
- c)3
- d)8
Correct answer is option 'B'. Can you explain this answer?
The number of rational terms in the expansion of
a)
6
b)
7
c)
3
d)
8
|
Vivek Kumar answered |
Use multinomial theorem.
General term of given expansion is
General term of given expansion is
23C0 + 23C2 + 23C4 + ... 23C22 equals - a)223 – 2
- b)222
- c)211
- d)
Correct answer is option 'B'. Can you explain this answer?
23C0 + 23C2 + 23C4 + ... 23C22 equals
a)
223 – 2
b)
222
c)
211
d)
|
|
Chirag Verma answered |
Given sum = sum of odd terms 
The product of middle terms in the expansion of 
is equal to - a)11C611C6
- b)
- c)11C611C6(x)
- d)(11C6)2x2
Correct answer is option 'A'. Can you explain this answer?
The product of middle terms in the expansion of is equal to
is equal to a)
11C611C6
b)
c)
11C611C6(x)
d)
(11C6)2x2
|
|
Chirag Verma answered |
it has 12 terms in it’s expansion , so there are two middle terms (6th and 7th); 
The sum of the binomial coefficients in the expansion of (x-3/4 + ax5/4)n lies between 200 and 400 and the term independent of x equals 448. The value of a is - a)1
- b)2
- c)1/2
- d)for no value of a
Correct answer is option 'B'. Can you explain this answer?
The sum of the binomial coefficients in the expansion of (x-3/4 + ax5/4)n lies between 200 and 400 and the term independent of x equals 448. The value of a is
a)
1
b)
2
c)
1/2
d)
for no value of a
|
Zara Khan answered |
Given:
The expansion of $(x-\frac{3}{4}ax^{\frac{5}{4}})^n$ has the following conditions:
1) The sum of the binomial coefficients is between 200 and 400.
2) The term independent of $x$ is 448.
To Find:
The value of $a$.
Solution:
1. Finding the Sum of Binomial Coefficients:
The sum of binomial coefficients in the expansion of $(x-\frac{3}{4}ax^{\frac{5}{4}})^n$ can be calculated using the binomial theorem.
The binomial theorem states that the $k^{th}$ term in the expansion of $(a+b)^n$ is given by:
$T_k = \binom{n}{k}a^{n-k}b^k$
In our case, $a = x$ and $b = -\frac{3}{4}ax^{\frac{5}{4}}$, and we need to find the sum of the binomial coefficients. Let's denote this sum as $S$.
$S = \binom{n}{0}a^n(-\frac{3}{4}ax^{\frac{5}{4}})^0 + \binom{n}{1}a^{n-1}(-\frac{3}{4}ax^{\frac{5}{4}})^1 + \binom{n}{2}a^{n-2}(-\frac{3}{4}ax^{\frac{5}{4}})^2 + ... + \binom{n}{n}(-\frac{3}{4}ax^{\frac{5}{4}})^n$
Since we are interested in the sum of binomial coefficients, we can ignore all terms with powers of $a$. This leaves us with:
$S = \binom{n}{0} + \binom{n}{1}(-\frac{3}{4}ax^{\frac{5}{4}}) + \binom{n}{2}(-\frac{3}{4}ax^{\frac{5}{4}})^2 + ... + \binom{n}{n}(-\frac{3}{4}ax^{\frac{5}{4}})^n$
Notice that this is the expansion of $(1 - \frac{3}{4}ax^{\frac{5}{4}})^n$.
So, $S = (1 - \frac{3}{4}ax^{\frac{5}{4}})^n$
Given that the sum of binomial coefficients is between 200 and 400, we have:
$200 < (1="" -="" \frac{3}{4}ax^{\frac{5}{4}})^n="" />< />
2. Finding the Term Independent of x:
The term independent of $x$ is obtained when all the powers of $x$ cancel out in the expansion. This occurs when the exponent of $x$ in each term is zero. In other words, the exponent of $x$ in each term is a multiple of $\frac{5}{4}$.
The term independent of $x$ in the expansion of $(x-\frac{3}{4}ax^{\frac{5}{4}})^
The expansion of $(x-\frac{3}{4}ax^{\frac{5}{4}})^n$ has the following conditions:
1) The sum of the binomial coefficients is between 200 and 400.
2) The term independent of $x$ is 448.
To Find:
The value of $a$.
Solution:
1. Finding the Sum of Binomial Coefficients:
The sum of binomial coefficients in the expansion of $(x-\frac{3}{4}ax^{\frac{5}{4}})^n$ can be calculated using the binomial theorem.
The binomial theorem states that the $k^{th}$ term in the expansion of $(a+b)^n$ is given by:
$T_k = \binom{n}{k}a^{n-k}b^k$
In our case, $a = x$ and $b = -\frac{3}{4}ax^{\frac{5}{4}}$, and we need to find the sum of the binomial coefficients. Let's denote this sum as $S$.
$S = \binom{n}{0}a^n(-\frac{3}{4}ax^{\frac{5}{4}})^0 + \binom{n}{1}a^{n-1}(-\frac{3}{4}ax^{\frac{5}{4}})^1 + \binom{n}{2}a^{n-2}(-\frac{3}{4}ax^{\frac{5}{4}})^2 + ... + \binom{n}{n}(-\frac{3}{4}ax^{\frac{5}{4}})^n$
Since we are interested in the sum of binomial coefficients, we can ignore all terms with powers of $a$. This leaves us with:
$S = \binom{n}{0} + \binom{n}{1}(-\frac{3}{4}ax^{\frac{5}{4}}) + \binom{n}{2}(-\frac{3}{4}ax^{\frac{5}{4}})^2 + ... + \binom{n}{n}(-\frac{3}{4}ax^{\frac{5}{4}})^n$
Notice that this is the expansion of $(1 - \frac{3}{4}ax^{\frac{5}{4}})^n$.
So, $S = (1 - \frac{3}{4}ax^{\frac{5}{4}})^n$
Given that the sum of binomial coefficients is between 200 and 400, we have:
$200 < (1="" -="" \frac{3}{4}ax^{\frac{5}{4}})^n="" />< />
2. Finding the Term Independent of x:
The term independent of $x$ is obtained when all the powers of $x$ cancel out in the expansion. This occurs when the exponent of $x$ in each term is zero. In other words, the exponent of $x$ in each term is a multiple of $\frac{5}{4}$.
The term independent of $x$ in the expansion of $(x-\frac{3}{4}ax^{\frac{5}{4}})^
The value of 1 + 1.1! + 2. 2! + 3.3! + …… + n. n! is - a)(n + 1)! + 1
- b)(n – 1)! + 1
- c)(n + 1)! – 1
- d)(n + 1)!
Correct answer is option 'D'. Can you explain this answer?
The value of 1 + 1.1! + 2. 2! + 3.3! + …… + n. n! is
a)
(n + 1)! + 1
b)
(n – 1)! + 1
c)
(n + 1)! – 1
d)
(n + 1)!
|
|
Chirag Verma answered |
n. n! = [(n + 1) – 1]n! = (n + 1)! – n!
∴ 1 + 2! – 1! + … + (n + 1)! – n! = (n + 1)!
∴ 1 + 2! – 1! + … + (n + 1)! – n! = (n + 1)!
The value of C1 + 3C3 + 5C5 + 7C7 + ...., where C0, C3, C5, C7,..... are binomial coefficients is - a)n.2n -1
- b)n.2n +1
- c)n.2n
- d)n.2n-2
Correct answer is option 'D'. Can you explain this answer?
The value of C1 + 3C3 + 5C5 + 7C7 + ...., where C0, C3, C5, C7,..... are binomial coefficients is
a)
n.2n -1
b)
n.2n +1
c)
n.2n
d)
n.2n-2
|
Jhanvi Agrawal answered |
The given expression involves binomial coefficients, which are calculated using the formula C(n, r) = n! / (r!(n-r)!), where n is the total number of items and r is the number of items chosen at a time.
To understand the given expression, let's break it down step by step:
Step 1: C1
C1 represents the binomial coefficient when n = 1 and r = 1. Plugging these values into the formula, we get C(1, 1) = 1! / (1!(1-1)!) = 1.
Step 2: 3C3
3C3 represents the binomial coefficient when n = 3 and r = 3. Plugging these values into the formula, we get C(3, 3) = 3! / (3!(3-3)!) = 1.
Step 3: 5C5
5C5 represents the binomial coefficient when n = 5 and r = 5. Plugging these values into the formula, we get C(5, 5) = 5! / (5!(5-5)!) = 1.
Step 4: 7C7
7C7 represents the binomial coefficient when n = 7 and r = 7. Plugging these values into the formula, we get C(7, 7) = 7! / (7!(7-7)!) = 1.
From the above steps, we can observe that the value of each term in the given expression is 1.
Now, let's generalize the pattern to find the value of the expression for any given value of n:
Step 5: C2n
C2n represents the binomial coefficient when n = 2n and r = 2n. Plugging these values into the formula, we get C(2n, 2n) = (2n)! / ((2n)!((2n)-(2n))!) = 1.
Therefore, for any value of n, the value of the expression C1 3C3 5C5 7C7 ... is always 1.
Hence, the correct answer is option 'D' - n.2n-2.
To understand the given expression, let's break it down step by step:
Step 1: C1
C1 represents the binomial coefficient when n = 1 and r = 1. Plugging these values into the formula, we get C(1, 1) = 1! / (1!(1-1)!) = 1.
Step 2: 3C3
3C3 represents the binomial coefficient when n = 3 and r = 3. Plugging these values into the formula, we get C(3, 3) = 3! / (3!(3-3)!) = 1.
Step 3: 5C5
5C5 represents the binomial coefficient when n = 5 and r = 5. Plugging these values into the formula, we get C(5, 5) = 5! / (5!(5-5)!) = 1.
Step 4: 7C7
7C7 represents the binomial coefficient when n = 7 and r = 7. Plugging these values into the formula, we get C(7, 7) = 7! / (7!(7-7)!) = 1.
From the above steps, we can observe that the value of each term in the given expression is 1.
Now, let's generalize the pattern to find the value of the expression for any given value of n:
Step 5: C2n
C2n represents the binomial coefficient when n = 2n and r = 2n. Plugging these values into the formula, we get C(2n, 2n) = (2n)! / ((2n)!((2n)-(2n))!) = 1.
Therefore, for any value of n, the value of the expression C1 3C3 5C5 7C7 ... is always 1.
Hence, the correct answer is option 'D' - n.2n-2.
f = R - [R], then Rf = 
The expansion [x + (x3 - 1)1/2]5 + [x + (x3 - 1)1/2]5 is a polynomial of degree- a)8
- b)7
- c)6
- d)5
Correct answer is option 'B'. Can you explain this answer?
The expansion [x + (x3 - 1)1/2]5 + [x + (x3 - 1)1/2]5 is a polynomial of degree
a)
8
b)
7
c)
6
d)
5
|
Shlok Shah answered |
Explanation:
To find the degree of the polynomial, we need to determine the highest power of x in the expression.
Expansion of the first term:
The expansion of [x(x^3 - 1)^1/2]^5 can be found by using the binomial theorem. According to the binomial theorem, the expansion of (a + b)^n is given by:
(a + b)^n = nC0 * a^n * b^0 + nC1 * a^(n-1) * b^1 + nC2 * a^(n-2) * b^2 + ... + nCn * a^0 * b^n
In our case, a = x, b = (x^3 - 1)^(1/2), and n = 5. Substituting these values, the expansion becomes:
[x(x^3 - 1)^1/2]^5 = 5C0 * x^5 * (x^3 - 1)^(1/2)^0 + 5C1 * x^4 * (x^3 - 1)^(1/2)^1 + 5C2 * x^3 * (x^3 - 1)^(1/2)^2 + ... + 5C5 * x^0 * (x^3 - 1)^(1/2)^5
Simplifying each term, we get:
1 * x^5 * 1 + 5 * x^4 * (x^3 - 1)^(1/2) + 10 * x^3 * (x^3 - 1) + ... + 5 * x * (x^3 - 1)^(1/2)^4 + 1 * (x^3 - 1)^(1/2)^5
Expansion of the second term:
Similarly, the expansion of [x(x^3 - 1)^1/2]^5 can be found by using the same process. Since it is the same expression, the expansion will be identical to the first term:
[x(x^3 - 1)^1/2]^5 = 1 * x^5 * 1 + 5 * x^4 * (x^3 - 1)^(1/2) + 10 * x^3 * (x^3 - 1) + ... + 5 * x * (x^3 - 1)^(1/2)^4 + 1 * (x^3 - 1)^(1/2)^5
Degree of the polynomial:
Since both terms have the same powers of x, we can add the corresponding coefficients of each power. The highest power of x will determine the degree of the polynomial.
The coefficients of the x^5 terms in both expansions are 1 * x^5 * 1 and 1 * x^5 * 1, which add up to 2x^5. Therefore, the highest power of x in the expansion is 5.
The degree of the polynomial is 5 (option D).
To find the degree of the polynomial, we need to determine the highest power of x in the expression.
Expansion of the first term:
The expansion of [x(x^3 - 1)^1/2]^5 can be found by using the binomial theorem. According to the binomial theorem, the expansion of (a + b)^n is given by:
(a + b)^n = nC0 * a^n * b^0 + nC1 * a^(n-1) * b^1 + nC2 * a^(n-2) * b^2 + ... + nCn * a^0 * b^n
In our case, a = x, b = (x^3 - 1)^(1/2), and n = 5. Substituting these values, the expansion becomes:
[x(x^3 - 1)^1/2]^5 = 5C0 * x^5 * (x^3 - 1)^(1/2)^0 + 5C1 * x^4 * (x^3 - 1)^(1/2)^1 + 5C2 * x^3 * (x^3 - 1)^(1/2)^2 + ... + 5C5 * x^0 * (x^3 - 1)^(1/2)^5
Simplifying each term, we get:
1 * x^5 * 1 + 5 * x^4 * (x^3 - 1)^(1/2) + 10 * x^3 * (x^3 - 1) + ... + 5 * x * (x^3 - 1)^(1/2)^4 + 1 * (x^3 - 1)^(1/2)^5
Expansion of the second term:
Similarly, the expansion of [x(x^3 - 1)^1/2]^5 can be found by using the same process. Since it is the same expression, the expansion will be identical to the first term:
[x(x^3 - 1)^1/2]^5 = 1 * x^5 * 1 + 5 * x^4 * (x^3 - 1)^(1/2) + 10 * x^3 * (x^3 - 1) + ... + 5 * x * (x^3 - 1)^(1/2)^4 + 1 * (x^3 - 1)^(1/2)^5
Degree of the polynomial:
Since both terms have the same powers of x, we can add the corresponding coefficients of each power. The highest power of x will determine the degree of the polynomial.
The coefficients of the x^5 terms in both expansions are 1 * x^5 * 1 and 1 * x^5 * 1, which add up to 2x^5. Therefore, the highest power of x in the expansion is 5.
The degree of the polynomial is 5 (option D).
The number of irrational terms in the expansion of (21/5 +31/10)55 is - a)47
- b)56
- c)50
- d)48
Correct answer is option 'C'. Can you explain this answer?
The number of irrational terms in the expansion of (21/5 +31/10)55 is
a)
47
b)
56
c)
50
d)
48
|
|
Chirag Verma answered |
(21/5 + 31/10)55
Total terms = 55 + 1 = 56
Here r = 0, 10, 20, 30, 40, 50
Number of rational terms = 6;
Number of irrational terms = 56 - 6 = 50
Total terms = 55 + 1 = 56
Here r = 0, 10, 20, 30, 40, 50
Number of rational terms = 6;
Number of irrational terms = 56 - 6 = 50
The number of terms in the expansion of (2x + 3y- 4z)n is - a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
The number of terms in the expansion of (2x + 3y- 4z)n is
a)
b)
c)
d)
|
|
Veda Institute answered |
We have, (2x + 3y - 4z)n = {2 + (3 - 4)}n
Clearly, the first term in the above expansion gives one term, second term gives two terms, third term gives three terms and so on.
So, Total number of term = 1 +2+3+...+n+(n+1) =
Clearly, the first term in the above expansion gives one term, second term gives two terms, third term gives three terms and so on.
So, Total number of term = 1 +2+3+...+n+(n+1) =
The number of irrational terms in the expansion of (21/5 + 31/10)55 is- a)47
- b)56
- c)50
- d)48
Correct answer is option 'C'. Can you explain this answer?
The number of irrational terms in the expansion of (21/5 + 31/10)55 is
a)
47
b)
56
c)
50
d)
48
|
|
Chirag Verma answered |
(21/5 + 31/10)55
Total terms = 55 + 1 = 56
Here r = 0, 10, 20, 30, 40, 50
Number of rational terms = 6;
Number of irrational terms = 56 ‐ 6 = 50
Total terms = 55 + 1 = 56
Here r = 0, 10, 20, 30, 40, 50
Number of rational terms = 6;
Number of irrational terms = 56 ‐ 6 = 50
The sum rCr + r+1Cr + r+2Cr + .... + nCr (n > r) equals - a)nCr+1
- b)n+1Cr+1
- c)n+1Cr-1
- d)n+1Cr
Correct answer is option 'B'. Can you explain this answer?
The sum rCr + r+1Cr + r+2Cr + .... + nCr (n > r) equals
a)
nCr+1
b)
n+1Cr+1
c)
n+1Cr-1
d)
n+1Cr
|
|
Chirag Verma answered |
C(n, r) + c(n ‐1, r) + C(n ‐ 2, r) + . . . + C(r, r)
= n+1Cr+1 (applying same rule again and again)
(∵ nCr + nCr‐1 = n+1Cr)
= n+1Cr+1 (applying same rule again and again)
(∵ nCr + nCr‐1 = n+1Cr)
If in the expansion of (1 + x)m(1 - x)n, the coefficient of x and x2 are 3 and -6 respectively, then m is - a)6
- b)9
- c)12
- d)24
Correct answer is option 'C'. Can you explain this answer?
If in the expansion of (1 + x)m(1 - x)n, the coefficient of x and x2 are 3 and -6 respectively, then m is
a)
6
b)
9
c)
12
d)
24
|
|
Chirag Verma answered |
⇒ m2 - m2 + n - n - 2mn = -12
⇒ (m - n)2 - (m + n) = -12 ⇒ m + n = 9 + 12 = 21 (2) using (1)
Solving (1) and (2), we get m = 12.
- a)2/9
- b)4/9
- c)2/3
- d)1/3
Correct answer is option 'A'. Can you explain this answer?
a)
2/9
b)
4/9
c)
2/3
d)
1/3
|
Bhaskar Yadav answered |
A will be the best answer
The product of middle terms in the expansion of 
is equal to- a)11C611C6
- b)
- c)11C611C6(x)
- d)(11C6)2x2
Correct answer is option 'A'. Can you explain this answer?
The product of middle terms in the expansion of is equal to
is equal toa)
11C611C6
b)
c)
11C611C6(x)
d)
(11C6)2x2
|
|
Veda Institute answered |
it has 12 terms in it’s expansion ,so there are two middle terms (6th and 7th);
The numbers of terms in the expansion of 
- a)201
- b)300
- c)200
- d)100C3
Correct answer is option 'A'. Can you explain this answer?
The numbers of terms in the expansion of
a)
201
b)
300
c)
200
d)
100C3
|
|
Veda Institute answered |
Where ao = sum of all absolute terms = 1 +100 C2 .2 + ....
Similarly a1, a2 , ...a100 and b1, b2 ...b100 are coefficients obtained after simplification.
∴ Total number of terms = 1 + 100 + 100 = 201
In the expansion of 
the coefficient of x-10 will be - a)12a11
- b)12b11a
- c)12a11b
- d)12a11b11
Correct answer is option 'C'. Can you explain this answer?
In the expansion of the coefficient of x-10 will be
the coefficient of x-10 will be a)
12a11
b)
12b11a
c)
12a11b
d)
12a11b11
|
|
Veda Institute answered |
Given expansion is 
∴ General term
Since, we have to find coefficient of x-10 ∴ -12 + 2r = -10 ⇒ r = 1
Now, then coefficient of x-10 is 12C1(a)11(b)1 = 12a11b
∴ General term
Since, we have to find coefficient of x-10 ∴ -12 + 2r = -10 ⇒ r = 1
Now, then coefficient of x-10 is 12C1(a)11(b)1 = 12a11b
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
a)
b)
c)
d)
|
|
Veda Institute answered |
∴ (d) is correct
In the expansion of 
the coefficient of x−10 will be - a)12a11
- b)12b11a
- c)12a11b
- d) 12a11b11
Correct answer is option 'C'. Can you explain this answer?
In the expansion of the coefficient of x−10 will be
the coefficient of x−10 will be a)
12a11
b)
12b11a
c)
12a11b
d)
12a11b11
|
|
Veda Institute answered |
Given expansion is 
Since, we have to find coefficient of x−10
∴ −12 + 2r = −10 ⇒ r = 1
Now, then coefficient of x−10 is 12C1(a)11(b)1 = 12a b
Since, we have to find coefficient of x−10
∴ −12 + 2r = −10 ⇒ r = 1
Now, then coefficient of x−10 is 12C1(a)11(b)1 = 12a b
The first integral term in the expansion of 
- a)2nd term
- b)3rd term
- c)4th term
- d)5th term
Correct answer is option 'C'. Can you explain this answer?
The first integral term in the expansion of
a)
2nd term
b)
3rd term
c)
4th term
d)
5th term
|
|
Chirag Verma answered |
For first integral term for r = 3;
If a, b, c ∈R+ form a A.P., then 
- a)A.P.
- b)G.P.
- c)H.P.
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
If a, b, c ∈R+ form a A.P., then
a)
A.P.
b)
G.P.
c)
H.P.
d)
none of these
|
|
Chirag Verma answered |
Since a, b, c are in A.P., therefore
The number of dissimilar terms in the expansion of (a + b + c)2n+1 – (a + b – c)2n+1 is - a)(n + 1)2
- b)(n – 1)2
- c)4n2 – 1
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The number of dissimilar terms in the expansion of (a + b + c)2n+1 – (a + b – c)2n+1 is
a)
(n + 1)2
b)
(n – 1)2
c)
4n2 – 1
d)
none of these
|
|
Chirag Verma answered |
Given expansion 
Number of dissimilar term = (2n + 1) + (2n -1) + ....(2n - 3) + .... + 5 + 3 + 1
Number of dissimilar term = (2n + 1) + (2n -1) + ....(2n - 3) + .... + 5 + 3 + 1
- a)2 x 69
- b)3 x 69
- c)610
- d)
Correct answer is option 'D'. Can you explain this answer?
a)
2 x 69
b)
3 x 69
c)
610
d)
|
|
Veda Institute answered |
Replacing x by -x,
Subtracting and putting
The number of irrational terms in the expansion of (21/5 + 31/10)55 is- a)47
- b)56
- c)50
- d)48
Correct answer is option 'C'. Can you explain this answer?
The number of irrational terms in the expansion of (21/5 + 31/10)55 is
a)
47
b)
56
c)
50
d)
48
|
|
Chirag Verma answered |
(21/5 + 31/10)55 Total terms = 55 + 1 = 56
Here r = 0, 10, 20, 30, 40, 50
Number of rational terms = 6;
Number of irrational terms = 56 - 6 = 50
Here r = 0, 10, 20, 30, 40, 50
Number of rational terms = 6;
Number of irrational terms = 56 - 6 = 50
The middle term in the expansion of (1 – 2x + x2)n is - a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
The middle term in the expansion of (1 – 2x + x2)n is
a)
b)
c)
d)
|
|
Chirag Verma answered |
Here 2n is even integer, therefore,
term will be the middle term.Now, (n + 1)th term in (1 - x)2n
where a, b, c are in AP and | a | < 1, | b | < 1, | c | < 1, then x, y, z are in 
The number of ways in which 7 Indians and 6 Pakistanis sit around a round table so that no two Indians are together is- a)(7!)2
- b)(6!)2
- c)6! 7!
- d)zero
Correct answer is option 'D'. Can you explain this answer?
The number of ways in which 7 Indians and 6 Pakistanis sit around a round table so that no two Indians are together is
a)
(7!)2
b)
(6!)2
c)
6! 7!
d)
zero
|
|
Chirag Verma answered |
In between 6 Pakistanis we have 6 gaps on a circular table, so 7 Indians cannot be arranged in 6 gaps.
Coefficient of t24 in (1 + t2)12 (1 + t12) (1 + t24) is- a)12C6 +3
- b)12C6 +1
- c)12C6
- d)12C6 +2
Correct answer is option 'D'. Can you explain this answer?
Coefficient of t24 in (1 + t2)12 (1 + t12) (1 + t24) is
a)
12C6 +3
b)
12C6 +1
c)
12C6
d)
12C6 +2
|
|
Chirag Verma answered |
coefficient of t24 is (12C12 +12C6 + 1)
⇒ coefficient of t24 is (2 +12C6) .
The letters of the word ‘ZENITH’ are permuted in all possible ways and the words thus formed are arranged as in a dictionary. The rank of the word ‘ZENITH’ is- a)616
- b)618
- c)597
- d)5930
Correct answer is option 'A'. Can you explain this answer?
The letters of the word ‘ZENITH’ are permuted in all possible ways and the words thus formed are arranged as in a dictionary. The rank of the word ‘ZENITH’ is
a)
616
b)
618
c)
597
d)
5930
|
|
Chirag Verma answered |
Rank is 5(5!) + 0.4! + 2(3!) + 1 x 2! + 1 x 1! + 1 = 616
The number of ways to arrange the letters of the word ‘GARDEN’ with vowels in alphabetical order is- a)360
- b)240
- c)120
- d)480
Correct answer is option 'A'. Can you explain this answer?
The number of ways to arrange the letters of the word ‘GARDEN’ with vowels in alphabetical order is
a)
360
b)
240
c)
120
d)
480
|
Vedang Reddy answered |
Understanding the Problem
To find the number of ways to arrange the letters of the word "GARDEN" with vowels in alphabetical order, we first identify the vowels and consonants in the word.
Vowels and Consonants
- Vowels: A, E
- Consonants: G, R, D, N
Step 1: Total Letters and Arrangements
The word "GARDEN" has 6 letters. If there were no restrictions, the total arrangements of these letters would be calculated as follows:
- Total arrangements = 6! = 720
Step 2: Arranging Vowels in Order
Since the vowels A and E must be in alphabetical order, we consider the arrangements where A comes before E.
Step 3: Fixing Vowels' Order
For every arrangement of the 6 letters, A and E can be placed in any of the 2 positions, but since they must be in alphabetical order, we only need to consider one arrangement (A before E).
Step 4: Choosing Positions for Vowels
We choose 2 out of the 6 positions for the vowels (A and E). The number of ways to choose 2 positions from 6 is given by:
- Number of ways to choose positions = C(6, 2) = 15
Step 5: Arranging Consonants
The remaining 4 positions will be filled by the consonants G, R, D, N. These can be arranged in:
- Arrangements of consonants = 4! = 24
Final Calculation
Now, we multiply the number of ways to choose the positions for the vowels by the arrangements of the consonants:
- Total arrangements with vowels in order = C(6, 2) * 4! = 15 * 24 = 360
Thus, the correct answer is option 'A' - 360.
To find the number of ways to arrange the letters of the word "GARDEN" with vowels in alphabetical order, we first identify the vowels and consonants in the word.
Vowels and Consonants
- Vowels: A, E
- Consonants: G, R, D, N
Step 1: Total Letters and Arrangements
The word "GARDEN" has 6 letters. If there were no restrictions, the total arrangements of these letters would be calculated as follows:
- Total arrangements = 6! = 720
Step 2: Arranging Vowels in Order
Since the vowels A and E must be in alphabetical order, we consider the arrangements where A comes before E.
Step 3: Fixing Vowels' Order
For every arrangement of the 6 letters, A and E can be placed in any of the 2 positions, but since they must be in alphabetical order, we only need to consider one arrangement (A before E).
Step 4: Choosing Positions for Vowels
We choose 2 out of the 6 positions for the vowels (A and E). The number of ways to choose 2 positions from 6 is given by:
- Number of ways to choose positions = C(6, 2) = 15
Step 5: Arranging Consonants
The remaining 4 positions will be filled by the consonants G, R, D, N. These can be arranged in:
- Arrangements of consonants = 4! = 24
Final Calculation
Now, we multiply the number of ways to choose the positions for the vowels by the arrangements of the consonants:
- Total arrangements with vowels in order = C(6, 2) * 4! = 15 * 24 = 360
Thus, the correct answer is option 'A' - 360.
Co + 3.C1 + 5.C2 + .... + (2n + 1).Cn =- a)(n + 1)2n
- b)(2n + 1)2n-1
- c)(2n +1)2n
- d)(n + 1)2n-1
Correct answer is option 'A'. Can you explain this answer?
Co + 3.C1 + 5.C2 + .... + (2n + 1).Cn =
a)
(n + 1)2n
b)
(2n + 1)2n-1
c)
(2n +1)2n
d)
(n + 1)2n-1
|
|
Chirag Verma answered |
Substitute ‘n’ and verify the options.
The coefficient of x50 in the expansion of
- a)1000C50
- b)1001C50
- c)1002C50
- d)21001
Correct answer is option 'C'. Can you explain this answer?
The coefficient of x50 in the expansion of
a)
1000C50
b)
1001C50
c)
1002C50
d)
21001
|
|
Veda Institute answered |
Subtract above equations,
Coefficient of x50 in S = coefficient of x50 in
C1 + 2C2.a + 3.C3.a2 + .....+ 2n.C2na2n-1 = - a)n(1 + a)n – 1
- b)n(1 + a)n
Correct answer is option ''. Can you explain this answer?
C1 + 2C2.a + 3.C3.a2 + .....+ 2n.C2na2n-1 =
a)
n(1 + a)n – 1
b)
n(1 + a)n
|
|
Veda Institute answered |
(1 + a)2n = Co + C1a + C2a2 + … + C2n a2n
Differentiate both sides w.r.t. ‘a’.
Differentiate both sides w.r.t. ‘a’.
Coefficient of x5 in the expansion of 
- a)30C5
- b)10C5
- c)20C5
- d)30C20
Correct answer is option 'A'. Can you explain this answer?
Coefficient of x5 in the expansion of
a)
30C5
b)
10C5
c)
20C5
d)
30C20
|
|
Veda Institute answered |
= Coefficient of x25 in (1 + x)30
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