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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
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.
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 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 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) =
f = R - [R], then Rf = 
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 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)
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); 
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.
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 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
| | Chirag Verma answered |
Use multinomial theorem.
General term of given expansion is
General term of given expansion is
The sum of coefficients of (1 + x − 3x2)2134 is - a)−1
- b)1
- c)0
- d)22134
Correct answer is option 'B'. Can you explain this answer?
The sum of coefficients of (1 + x − 3x2)2134 is
a)
−1
b)
1
c)
0
d)
22134
| | Chirag Verma answered |
For the sum of coefficient, put x = 1, to obtain the sum is (1 + 1 ‐ 3)2134 = 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
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) .
2nC2 + 2nC4 + .......+ 2nC2n = - a)22n
- b)22n -1
- c)22n-1
- d)22n-1 -1
Correct answer is option 'D'. Can you explain this answer?
2nC2 + 2nC4 + .......+ 2nC2n =
a)
22n
b)
22n -1
c)
22n-1
d)
22n-1 -1
| | Chirag Verma answered |
Substitute a positive integer for ‘n’ and verify.
Coefficient of x50 in (1 + x)1000 + 2x(1 + x)999 + 3x(1 + x)998 + .....is- a)1000C50
- b)1001C50
- c)1002C50
- d)1002C49
Correct answer is option 'C'. Can you explain this answer?
Coefficient of x50 in (1 + x)1000 + 2x(1 + x)999 + 3x(1 + x)998 + .....is
a)
1000C50
b)
1001C50
c)
1002C50
d)
1002C49
| | Chirag Verma answered |
Take (1 + x)1000 as common, after simplification it becomes (1 + x)1000 coefficient of x50 is 1002C50
The term independent of x in 
- a)5
- b)7
- c)6
- d)8
Correct answer is option 'C'. Can you explain this answer?
The term independent of x in
a)
5
b)
7
c)
6
d)
8
| | Chirag Verma answered |
The general term 
The term independent of x, (or the constant term) corresponds to x18-3r being x0 or 18 - 3r = 0 ⇒ r = 6 .
The term independent of x, (or the constant term) corresponds to x18-3r being x0 or 18 - 3r = 0 ⇒ r = 6 .
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
a)
b)
c)
d)
| | Chirag Verma answered |
∴ (d) is correct
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
- 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)
| | Chirag Verma answered |
Replacing x by -x,
Subtracting and putting
upto n terms 
Coefficient of x4 in (1 + x – 2x2)6 is- a)–60
- b)–45
- c)45
- d)15
Correct answer is option 'B'. Can you explain this answer?
Coefficient of x4 in (1 + x – 2x2)6 is
a)
–60
b)
–45
c)
45
d)
15
 | Ayush Mehra answered |
Expanding the expression
To find the coefficient of \(x^4\) in the given expression \((1 + x – 2x^2)^6\), we need to expand the expression using the binomial theorem.
Binomial theorem
The binomial theorem states that for any real numbers a and b, and any non-negative integer n, the expansion of \((a + b)^n\) can be written as the sum of terms of the form \({n \choose k} a^{n-k} b^k\), where \({n \choose k}\) is the binomial coefficient.
Coefficient of x^4
To find the coefficient of \(x^4\) in the expansion, we need to look for terms of the form \(x^4\) in the expanded expression. This can be achieved by finding all possible combinations of terms from the base expression \((1 + x – 2x^2)\) that multiply to give \(x^4\).
Calculating the coefficient
When expanding the given expression, the term that contributes to \(x^4\) is \({6 \choose 2} \cdot 1^4 \cdot x^4 \cdot (-2x^2)^2\). Calculating this term gives \({6 \choose 2} \cdot 1^4 \cdot x^4 \cdot 4x^4 = 15 \cdot 1 \cdot 4 = 60\).
Therefore, the coefficient of \(x^4\) in the expression \((1 + x – 2x^2)^6\) is 60.
So, option B (-45) is not correct.
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 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)
= r+1Cr+1 + r+1Cr + r+2Cr + .... + n-1Cr + nCr
= n+1Cr+1 (applying same rule again and again ) (∴ nCr + nCr-1 = n+1Cr)
= r+1Cr+1 + r+1Cr + r+2Cr + .... + n-1Cr + nCr
= n+1Cr+1 (applying same rule again and again ) (∴ nCr + nCr-1 = n+1Cr)
= I + F when I is odd and 0 < F < 1, then (I + F) (I - F) =
The coefficient of the term independent of x in the expansion of 
- a)210
- b)105
- c)70
- d)112
Correct answer is option 'A'. Can you explain this answer?
The coefficient of the term independent of x in the expansion of
a)
210
b)
105
c)
70
d)
112
| | Chirag Verma answered |
We
for this term to be independent of x. we must have
So, the required coefficient is
10C4(-1)4 = 210
for this term to be independent of x. we must have
So, the required coefficient is
10C4(-1)4 = 210
If 22006 - 2006 divided by 7, the remainder is- a)0
- b)1
- c)2
- d)4
Correct answer is option 'A'. Can you explain this answer?
If 22006 - 2006 divided by 7, the remainder is
a)
0
b)
1
c)
2
d)
4
 | Gaurav Rao answered |
Problem:
Find the remainder when 22006 - 2006 is divided by 7.
Solution:
To find the remainder when a number is divided by another number, we can use the concept of congruence. Two numbers are said to be congruent modulo n if their difference is divisible by n.
Let's represent the given expression as a congruence equation:
22006 - 2006 ≡ x (mod 7)
Step 1: Simplify the expression on the left side of the congruence equation.
22006 - 2006 = 20000
The congruence equation becomes:
20000 ≡ x (mod 7)
Step 2: Find the remainder when 20000 is divided by 7.
To find the remainder, we divide 20000 by 7 and observe the remainder.
20000 ÷ 7 = 2857 remainder 1
Therefore, 20000 leaves a remainder of 1 when divided by 7.
Step 3: Substitute the remainder in the congruence equation.
x ≡ 1 (mod 7)
Answer:
The remainder when 22006 - 2006 is divided by 7 is 1. Therefore, option 'A' is the correct answer.
Find the remainder when 22006 - 2006 is divided by 7.
Solution:
To find the remainder when a number is divided by another number, we can use the concept of congruence. Two numbers are said to be congruent modulo n if their difference is divisible by n.
Let's represent the given expression as a congruence equation:
22006 - 2006 ≡ x (mod 7)
Step 1: Simplify the expression on the left side of the congruence equation.
22006 - 2006 = 20000
The congruence equation becomes:
20000 ≡ x (mod 7)
Step 2: Find the remainder when 20000 is divided by 7.
To find the remainder, we divide 20000 by 7 and observe the remainder.
20000 ÷ 7 = 2857 remainder 1
Therefore, 20000 leaves a remainder of 1 when divided by 7.
Step 3: Substitute the remainder in the congruence equation.
x ≡ 1 (mod 7)
Answer:
The remainder when 22006 - 2006 is divided by 7 is 1. Therefore, option 'A' is the correct answer.
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
| | Chirag Verma 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
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
| | Chirag Verma answered |
Subtract above equations,
Coefficient of x50 in S = coefficient of x50 in
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