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All questions of Binomial Theorem & its Simple Applications for JEE Exam
In the expansion of the binomial expansion (a + b)n, which of the following is incorrect ?
- a)The exponent of a decreases from n to 0.
- b)The number of terms is n+1.
- c)The exponent of b increases from 0 to n.
- d)The binomial coefficients in (a+b)n, which are equidistant from beginning and end are not equal.
Correct answer is option 'D'. Can you explain this answer?
In the expansion of the binomial expansion (a + b)n, which of the following is incorrect ?
a)
The exponent of a decreases from n to 0.
b)
The number of terms is n+1.
c)
The exponent of b increases from 0 to n.
d)
The binomial coefficients in (a+b)n, which are equidistant from beginning and end are not equal.
|
Krishna Iyer answered |
Correct Answer: d
Explanation:- The coefficient of terms (x+a)n equidistant from the beginning and the end are equal. These coefficients are known as the binomial coefficients.
nCr = nCn – r, r = 0,1,2,…,n.
The middle term in the expansion of (x + y)10 is the- a)6th term
- b)5th term
- c)average of 5th and 6th terms
- d)7th term
Correct answer is option 'A'. Can you explain this answer?
The middle term in the expansion of (x + y)10 is the
a)
6th term
b)
5th term
c)
average of 5th and 6th terms
d)
7th term
|
Vikas Kapoor answered |
Number of terms(n) = 10
Middle term = (n/2) + 1
Middle term = (n/2) + 1
= (10/2) + 1
= 5 + 1
= 6th term
= 5 + 1
= 6th term
is
The number of terms in the expansion of (2x - 3y)8 is- a)6
- b)9
- c)8
- d)5
Correct answer is option 'B'. Can you explain this answer?
The number of terms in the expansion of (2x - 3y)8 is
a)
6
b)
9
c)
8
d)
5
|
Aryan Khanna answered |
Since this binomial is to the power 8, there will be nine terms in the expansion.
The number of terms in the expansion of (x – y + 2z)7 are:- a)35
- b)38
- c)36
- d)37
Correct answer is option 'C'. Can you explain this answer?
The number of terms in the expansion of (x – y + 2z)7 are:
a)
35
b)
38
c)
36
d)
37
|
Tejas Verma answered |
Here the number of terms can be calculated by:
= ((n+ 1) * (n+2)) /2
where, n =7
= ((n+ 1) * (n+2)) /2
where, n =7
∴ Number of terms = 36
are equal, then n is equal to
In the expansion of (a+b)n, nN the number of terms is:- a)n + 1
- b)n
- c)n – 1
- d)an
Correct answer is option 'A'. Can you explain this answer?
In the expansion of (a+b)n, nN the number of terms is:
a)
n + 1
b)
n
c)
n – 1
d)
an
|
Sruthi Manoj answered |
For example if it is (a b)², then the expansion is a² 2ab b² which consist of 3 terms. Similarly (a b)¹= a b which consists of 2 terms. therefore that implies (a b) raised to n, then the no. of terms is n 1.
If n is a +ve integer, then the binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are- a)additive inverse of each other
- b)multiplicative inverse of each other
- c)equal
- d)nothing can be said
Correct answer is option 'C'. Can you explain this answer?
If n is a +ve integer, then the binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are
a)
additive inverse of each other
b)
multiplicative inverse of each other
c)
equal
d)
nothing can be said
|
Hansa Sharma answered |
(x+a)n = nC0 xn + nC1 x(n-1) a1 + nC2 x(n-2) a2 + ..........+ nC(n-1) xa(n-1) + nCn an
Now, nC0 = nCn, nC1 = nCn-1, nC2 = nCn-2,........
therefore, nCr = nCn-r
The binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are equal.
Now, nC0 = nCn, nC1 = nCn-1, nC2 = nCn-2,........
therefore, nCr = nCn-r
The binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are equal.
The expansion of 
, in powers of x, is valid if- a)|x| > 2
- b)x < 2
- c)x > 2
- d)|x| < 2
Correct answer is option 'D'. Can you explain this answer?
The expansion of , in powers of x, is valid if
, in powers of x, is valid ifa)
|x| > 2
b)
x < 2
c)
x > 2
d)
|x| < 2
|
Knowledge Hub answered |
In case of negative or fractional power, expansion (1+x)^n is valid only when |x| < 1
(6 - 3x)-1/2
= (6-1/2 (1 - x/2)-1/2)
So, this equation exists only when |x/2| < 1
|x| < 2
(6 - 3x)-1/2
= (6-1/2 (1 - x/2)-1/2)
So, this equation exists only when |x/2| < 1
|x| < 2
The coefficient of x3 in the binomial expansion of 
- a)792m5
- b)942m7
- c)330m4
- d)792m6
Correct answer is option 'C'. Can you explain this answer?
The coefficient of x3 in the binomial expansion of
a)
792m5
b)
942m7
c)
330m4
d)
792m6
|
Geetika Shah answered |
T(r+1) = 11Cr x(11-2r) (m/r)r (-1)r
= 11Cr x(11-2r) mr (-1)r
Coefficient of x3 = 11 - 2r = 3
8 = 2r
r = 4
T5 = 11C4 x3 m4 (-1)
Coefficient of x3 = 11C4 m4
= 330 m4
= 11Cr x(11-2r) mr (-1)r
Coefficient of x3 = 11 - 2r = 3
8 = 2r
r = 4
T5 = 11C4 x3 m4 (-1)
Coefficient of x3 = 11C4 m4
= 330 m4
In the expansion of (1+x)60, the sum of coefficients of odd powers of x is- a)261
- b)260
- c)259
- d)none of these.
Correct answer is option 'C'. Can you explain this answer?
In the expansion of (1+x)60, the sum of coefficients of odd powers of x is
a)
261
b)
260
c)
2
59
d)
none of these.
|
|
Mohit Mittal answered |
C is correct as
sum of ( 1+ x )^60 gives 2^60 which includes both even and odd powers of x
so for only one type of power ( odd power) of x we divide 2^ 60 by 2 so we get 2^ 59
sum of ( 1+ x )^60 gives 2^60 which includes both even and odd powers of x
so for only one type of power ( odd power) of x we divide 2^ 60 by 2 so we get 2^ 59
The general term in the expansion of (a - b)n is- a)Tr+1 = (-1)r nCr an-r br
- b)Tr+1 = nCr an-r br
- c)Tr+1 = nCr ar br
- d)Tr+1 = (-1)r nCr ar br
Correct answer is option 'A'. Can you explain this answer?
The general term in the expansion of (a - b)n is
a)
Tr+1 = (-1)r nCr an-r br
b)
Tr+1 = nCr an-r br
c)
Tr+1 = nCr ar br
d)
Tr+1 = (-1)r nCr ar br
|
Poonam Reddy answered |
If a and b are real numbers and n is a positive integer, then:
(a - b)n = nC0 an + nC1 a(n – 1) b1 + nC2 a(n – 2) b2+ ...... + nCr a(n – r) br+ ... + nCnbn,
(a - b)n = nC0 an + nC1 a(n – 1) b1 + nC2 a(n – 2) b2+ ...... + nCr a(n – r) br+ ... + nCnbn,
The general term or (r + 1)th term in the expansion is given by:
Tr + 1 = (-1)Cr a(n–r) br
Tr + 1 = (-1)Cr a(n–r) br
What is the coefficient of x5 in the expansion of (1-x)-6 ?- a)252
- b)250
- c)-252
- d)251
Correct answer is option 'A'. Can you explain this answer?
What is the coefficient of x5 in the expansion of (1-x)-6 ?
a)
252
b)
250
c)
-252
d)
251
|
Preeti Iyer answered |
(1-x)-6
=> (1-x)(-6/1)
It is in the form of (1-x)(-p/q), p =6, q=1
=> (1-x)(-6/1)
It is in the form of (1-x)(-p/q), p =6, q=1
(1-x)(-p/q) = 1+p/1!(x/q)1 + p(p+q)/2!(x/q)2 + p(p+q)(p+2q)/3!(x/q)3 + p(p+q)(p+2q)(p+3q)/4!(x/q)4........
= 1+6/1!(x/1)1 + 6(7)/2!(x/1)2 + 6(7)(8)/3!(x/1)3 + 6(7)(8)(9)/4!(x/1)4 +.......................
So, coefficient of x5 is (6*7*8*9*10)/120
= 252
= 252
If in the expansion of (1+x)20, the coefficients of rth and (r+4)th terms are equal, then the value of r is equal to:- a)9
- b)7
- c)10
- d)8
Correct answer is option 'A'. Can you explain this answer?
If in the expansion of (1+x)20, the coefficients of rth and (r+4)th terms are equal, then the value of r is equal to:
a)
9
b)
7
c)
10
d)
8
|
Raghav Bansal answered |
Coefficients of the rth and (r+4)th terms in the given expansion are Cr−120 and 20Cr+3.
Here,Cr−120 = 20Cr+3
⇒ r−1+r+3 = 20
[∵ if nCx = nCy ⇒ x = y or x+y = n]
Here,Cr−120 = 20Cr+3
⇒ r−1+r+3 = 20
[∵ if nCx = nCy ⇒ x = y or x+y = n]
⇒ r = 2 or 2r = 18
⇒ r = 9
⇒ r = 9
The coefficient of y in the expansion of (y² + c/y)5 is - a)10c
- b)10c²
- c)10c³
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
The coefficient of y in the expansion of (y² + c/y)5 is
a)
10c
b)
10c²
c)
10c³
d)
None of these
|
|
Varun Kapoor answered |
Given, binomial expression is (y² + c / y)5
Now, Tr+1 = 5Cr × (y²)5-r × (c / y)r
= 5Cr × y10-3r × Cr
Now, 10 – 3r = 1
⇒ 3r = 9
⇒ r = 3
So, the coefficient of y = 5C3 × c³ = 10c³
Now, Tr+1 = 5Cr × (y²)5-r × (c / y)r
= 5Cr × y10-3r × Cr
Now, 10 – 3r = 1
⇒ 3r = 9
⇒ r = 3
So, the coefficient of y = 5C3 × c³ = 10c³
The coefficient of x17 in the expansion of (x- 1) (x- 2) …..(x – 18) is- a)- 171
- b)342
- c)171/2
- d)684
Correct answer is option 'A'. Can you explain this answer?
The coefficient of x17 in the expansion of (x- 1) (x- 2) …..(x – 18) is
a)
- 171
b)
342
c)
171/2
d)
684
|
Infinity Academy answered |
The coefficient of x17 is given by
−1 + (−2) + (−3) + ….. (−18)
= −1 − 2 − 3….. − 18
= − (18(18+1))/2
= − 9(19)
= − 171
−1 + (−2) + (−3) + ….. (−18)
= −1 − 2 − 3….. − 18
= − (18(18+1))/2
= − 9(19)
= − 171
If the expansion of 
in powers of x contains the term x2r, then n−2r is- a)even
- b)odd
- c)a positive integral multiple of 5
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
If the expansion of in powers of x contains the term x2r, then n−2r is
in powers of x contains the term x2r, then n−2r isa)
even
b)
odd
c)
a positive integral multiple of 5
d)
none of these
|
|
Geetika Shah answered |
T(r+1) = nCr xn-r ar
T(r+1) = n-5Cr xn-5-r (1/x4)r
= n-5Cr xn-5-r (1/x4r)
= n-5Cr xn-5-5r
=> n - 5 - 5r = 2r
=> n - 2r = 5(r + 1)
T(r+1) = n-5Cr xn-5-r (1/x4)r
= n-5Cr xn-5-r (1/x4r)
= n-5Cr xn-5-5r
=> n - 5 - 5r = 2r
=> n - 2r = 5(r + 1)
The greatest coefficient in the expansion of (1+x)12 is- a)C (12, 4)
- b)C (12, 6)
- c)C (12, 5)
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
The greatest coefficient in the expansion of (1+x)12 is
a)
C (12, 4)
b)
C (12, 6)
c)
C (12, 5)
d)
none of these
|
|
Harshad Unni answered |
Explanation:
To find the greatest coefficient in the expansion of (1 + x)^12, we can use the binomial theorem. The binomial theorem states that for any positive integer n, the expansion of (a + b)^n can be written as:
(a + b)^n = C(n, 0) * a^n * b^0 + C(n, 1) * a^(n-1) * b^1 + C(n, 2) * a^(n-2) * b^2 + ... + C(n, n-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n
where C(n, k) represents the binomial coefficient, also known as "n choose k".
In this case, we have (1 + x)^12, so a = 1 and b = x. Plugging these values into the binomial theorem formula, we get:
(1 + x)^12 = C(12, 0) * 1^12 * x^0 + C(12, 1) * 1^11 * x^1 + C(12, 2) * 1^10 * x^2 + ... + C(12, 11) * 1^1 * x^11 + C(12, 12) * 1^0 * x^12
Simplifying this expression, we have:
(1 + x)^12 = C(12, 0) + C(12, 1) * x + C(12, 2) * x^2 + ... + C(12, 11) * x^11 + C(12, 12) * x^12
To find the greatest coefficient, we need to determine which term has the highest coefficient. The coefficient of each term is given by the binomial coefficient C(12, k), where k represents the power of x.
Comparing Coefficients:
In order to find the greatest coefficient, we need to compare the binomial coefficients for each term. The binomial coefficient C(12, k) can be calculated using the formula:
C(n, k) = n! / (k! * (n - k)!)
where "!" represents the factorial operation.
Let's calculate the binomial coefficients for the given options:
a) C(12, 4) = 12! / (4! * (12 - 4)!) = 12! / (4! * 8!) = 495
b) C(12, 6) = 12! / (6! * (12 - 6)!) = 12! / (6! * 6!) = 924
c) C(12, 5) = 12! / (5! * (12 - 5)!) = 12! / (5! * 7!) = 792
Comparing the coefficients, we can see that option B has the greatest coefficient, which is 924. Therefore, the correct answer is option B) C(12, 6).
To find the greatest coefficient in the expansion of (1 + x)^12, we can use the binomial theorem. The binomial theorem states that for any positive integer n, the expansion of (a + b)^n can be written as:
(a + b)^n = C(n, 0) * a^n * b^0 + C(n, 1) * a^(n-1) * b^1 + C(n, 2) * a^(n-2) * b^2 + ... + C(n, n-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n
where C(n, k) represents the binomial coefficient, also known as "n choose k".
In this case, we have (1 + x)^12, so a = 1 and b = x. Plugging these values into the binomial theorem formula, we get:
(1 + x)^12 = C(12, 0) * 1^12 * x^0 + C(12, 1) * 1^11 * x^1 + C(12, 2) * 1^10 * x^2 + ... + C(12, 11) * 1^1 * x^11 + C(12, 12) * 1^0 * x^12
Simplifying this expression, we have:
(1 + x)^12 = C(12, 0) + C(12, 1) * x + C(12, 2) * x^2 + ... + C(12, 11) * x^11 + C(12, 12) * x^12
To find the greatest coefficient, we need to determine which term has the highest coefficient. The coefficient of each term is given by the binomial coefficient C(12, k), where k represents the power of x.
Comparing Coefficients:
In order to find the greatest coefficient, we need to compare the binomial coefficients for each term. The binomial coefficient C(12, k) can be calculated using the formula:
C(n, k) = n! / (k! * (n - k)!)
where "!" represents the factorial operation.
Let's calculate the binomial coefficients for the given options:
a) C(12, 4) = 12! / (4! * (12 - 4)!) = 12! / (4! * 8!) = 495
b) C(12, 6) = 12! / (6! * (12 - 6)!) = 12! / (6! * 6!) = 924
c) C(12, 5) = 12! / (5! * (12 - 5)!) = 12! / (5! * 7!) = 792
Comparing the coefficients, we can see that option B has the greatest coefficient, which is 924. Therefore, the correct answer is option B) C(12, 6).
The sum of coefficients in the expansion of (x+2y+z)n is (n being a positive integer)- a)3n
- b)2n
- c)1
- d)none of these
Correct answer is option 'D'. Can you explain this answer?
The sum of coefficients in the expansion of (x+2y+z)n is (n being a positive integer)
a)
3n
b)
2n
c)
1
d)
none of these
|
Maitri Kumar answered |
Explanation:
To find the sum of coefficients in the expansion of (x + 2y + z)^n, we can use the Binomial Theorem.
The Binomial Theorem states that for any positive integer n, the expansion of (x + y)^n can be written as:
(x + y)^n = C(n, 0)x^n y^0 + C(n, 1)x^(n-1) y^1 + C(n, 2)x^(n-2) y^2 + ... + C(n, n-1)x^1 y^(n-1) + C(n, n)x^0 y^n
Where C(n, k) is the binomial coefficient, given by:
C(n, k) = n! / (k!(n-k)!)
In the case of (x + 2y + z)^n, the coefficients will be found by replacing x with 1, y with 2, and z with 1. Therefore, the expansion becomes:
(1 + 2(2y) + 1)^n = (1 + 4y + 1)^n
Now we can apply the Binomial Theorem to find the coefficients. The sum of the coefficients will be the sum of the terms in the expansion.
Applying the Binomial Theorem:
Using the Binomial Theorem, the expansion of (1 + 4y + 1)^n can be written as:
(1 + 4y + 1)^n = C(n, 0)(1)^n (4y)^0 + C(n, 1)(1)^(n-1) (4y)^1 + C(n, 2)(1)^(n-2) (4y)^2 + ... + C(n, n-1)(1)^1 (4y)^(n-1) + C(n, n)(1)^0 (4y)^n
Simplifying this expression, we get:
(1 + 4y + 1)^n = C(n, 0) + 4C(n, 1)y + 6C(n, 2)(4y)^2 + ... + nC(n, n-1)(4y)^(n-1) + C(n, n)(4y)^n
The sum of the coefficients will be the sum of the terms in this expansion.
Calculating the Sum of the Coefficients:
To calculate the sum of the coefficients, we need to sum up all the terms in the expansion.
The sum of the coefficients can be written as:
C(n, 0) + 4C(n, 1) + 6C(n, 2) + ... + nC(n, n-1) + C(n, n)
Each term in this sum is a binomial coefficient multiplied by a constant factor.
Conclusion:
In general, the sum of the coefficients in the expansion of (x + 2y + z)^n is not equal to any of the given options (a, b, c). Therefore, the correct answer is option D, "none of these".
To find the sum of coefficients in the expansion of (x + 2y + z)^n, we can use the Binomial Theorem.
The Binomial Theorem states that for any positive integer n, the expansion of (x + y)^n can be written as:
(x + y)^n = C(n, 0)x^n y^0 + C(n, 1)x^(n-1) y^1 + C(n, 2)x^(n-2) y^2 + ... + C(n, n-1)x^1 y^(n-1) + C(n, n)x^0 y^n
Where C(n, k) is the binomial coefficient, given by:
C(n, k) = n! / (k!(n-k)!)
In the case of (x + 2y + z)^n, the coefficients will be found by replacing x with 1, y with 2, and z with 1. Therefore, the expansion becomes:
(1 + 2(2y) + 1)^n = (1 + 4y + 1)^n
Now we can apply the Binomial Theorem to find the coefficients. The sum of the coefficients will be the sum of the terms in the expansion.
Applying the Binomial Theorem:
Using the Binomial Theorem, the expansion of (1 + 4y + 1)^n can be written as:
(1 + 4y + 1)^n = C(n, 0)(1)^n (4y)^0 + C(n, 1)(1)^(n-1) (4y)^1 + C(n, 2)(1)^(n-2) (4y)^2 + ... + C(n, n-1)(1)^1 (4y)^(n-1) + C(n, n)(1)^0 (4y)^n
Simplifying this expression, we get:
(1 + 4y + 1)^n = C(n, 0) + 4C(n, 1)y + 6C(n, 2)(4y)^2 + ... + nC(n, n-1)(4y)^(n-1) + C(n, n)(4y)^n
The sum of the coefficients will be the sum of the terms in this expansion.
Calculating the Sum of the Coefficients:
To calculate the sum of the coefficients, we need to sum up all the terms in the expansion.
The sum of the coefficients can be written as:
C(n, 0) + 4C(n, 1) + 6C(n, 2) + ... + nC(n, n-1) + C(n, n)
Each term in this sum is a binomial coefficient multiplied by a constant factor.
Conclusion:
In general, the sum of the coefficients in the expansion of (x + 2y + z)^n is not equal to any of the given options (a, b, c). Therefore, the correct answer is option D, "none of these".
In a binomial expansion with power 13- a)one of the middle term is the 7th term
- b)the middle term is the average of the 6th and the 7th terms.
- c)the middle term is obtained by subtracting the 6th term from the 7th term.
- d)there are two middle terms, 6th and 7th
Correct answer is option 'A'. Can you explain this answer?
In a binomial expansion with power 13
a)
one of the middle term is the 7th term
b)
the middle term is the average of the 6th and the 7th terms.
c)
the middle term is obtained by subtracting the 6th term from the 7th term.
d)
there are two middle terms, 6th and 7th
|
Lavanya Menon answered |
is
5th term from the end in the expansion of 
- a)7920x4
- b)−7920x4
- c)7920x−4
- d)−7920x−4
Correct answer is option 'C'. Can you explain this answer?
5th term from the end in the expansion of
a)
7920x4
b)
−7920x4
c)
7920x−4
d)
−7920x−4
|
Akanksha Reddy answered |
5th term in the expansion of (x2/2 − 2/x2)12 is
Tr+ 1= nCr xr* y(n−r)
T5 = 12C4[(x2/2)4 (-2/x2)8]
= (12! * x8 * 28)/ (4! * 8! *24 * x16)
= 7920x−4
Tr+ 1= nCr xr* y(n−r)
T5 = 12C4[(x2/2)4 (-2/x2)8]
= (12! * x8 * 28)/ (4! * 8! *24 * x16)
= 7920x−4
The coefficient of xn in expansion of (1+x)(1−x)n is- a)(−1)n(n+1)
- b)(−1)n(1−n)
- c)(−1)n
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
The coefficient of xn in expansion of (1+x)(1−x)n is
a)
(−1)n(n+1)
b)
(−1)n(1−n)
c)
(−1)n
d)
none of these
|
|
Lavanya Menon answered |
We can expand (1+x)(1−x)n as
= (1+x)(nC0− nC1.x + nC2.x2 + ........ +(−1)n nCnxn)
= (1+x)(nC0− nC1.x + nC2.x2 + ........ +(−1)n nCnxn)
Coefficient of xn is
(−1)n nCn + (−1)n−1 nCn−1
=(−1)n(1−n)
If x = 9950+1005 and y = (101)50, then- a)x < y
- b)x = y
- c)x > y
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
If x = 9950+1005 and y = (101)50, then
a)
x < y
b)
x = y
c)
x > y
d)
none of these
|
|
Nitya Yadav answered |
The given equations are:
Ifx = 9950
1005andy = (101)50
To solve for x, we divide both sides of the first equation by 100:
Ifx/100 = 9950/100
x = 99.5
To solve for y, we divide both sides of the second equation by 1005:
andy = (101)50/1005
andy = 50
Therefore, the values of x and y are:
x = 99.5
y = 50
Ifx = 9950
1005andy = (101)50
To solve for x, we divide both sides of the first equation by 100:
Ifx/100 = 9950/100
x = 99.5
To solve for y, we divide both sides of the second equation by 1005:
andy = (101)50/1005
andy = 50
Therefore, the values of x and y are:
x = 99.5
y = 50
The coefficients of xn in the expansion of (1+2x + 3x2 + ........)1/2 is - a)1
- b)n
- c)n+1
- d)-1
Correct answer is option 'A'. Can you explain this answer?
The coefficients of xn in the expansion of (1+2x + 3x2 + ........)1/2 is
a)
1
b)
n
c)
n+1
d)
-1
|
|
Aarya Rane answered |
To find the coefficients of xn in the expansion of (1 + 2x + 3x^2 + ...)^1/2, we can use the binomial theorem. The binomial theorem states that for any real number a and b and any positive integer n, the expansion of (a + b)^n can be written as:
(a + b)^n = C(n, 0) * a^n * b^0 + C(n, 1) * a^(n-1) * b^1 + C(n, 2) * a^(n-2) * b^2 + ... + C(n, n-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n
where C(n, r) is the binomial coefficient, also called "n choose r", and is given by the formula:
C(n, r) = n! / (r! * (n-r)!)
In the given expression (1 + 2x + 3x^2 + ...)^1/2, we have a = 1 and b = 2x + 3x^2 + ... Therefore, applying the binomial theorem, the expansion becomes:
(1 + 2x + 3x^2 + ...)^1/2 = C(1/2, 0) * 1^(1/2) * (2x + 3x^2 + ...)^0 + C(1/2, 1) * 1^(1/2 - 1) * (2x + 3x^2 + ...)^1/2 + ...
Simplifying this expression, we find that the coefficient of xn is given by:
C(1/2, n) * 1^(1/2 - n) * (2x + 3x^2 + ...)^n/2
Since C(1/2, n) = 1 for all values of n, and 1^(1/2 - n) = 1, the coefficient of xn simplifies to:
(2x + 3x^2 + ...)^n/2
From this, we can see that the coefficient of xn is simply the expression (2x + 3x^2 + ...)^n/2. Since this expression does not depend on n, the coefficient of xn is constant and equal to 1.
Therefore, the correct answer is option A) 1.
(a + b)^n = C(n, 0) * a^n * b^0 + C(n, 1) * a^(n-1) * b^1 + C(n, 2) * a^(n-2) * b^2 + ... + C(n, n-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n
where C(n, r) is the binomial coefficient, also called "n choose r", and is given by the formula:
C(n, r) = n! / (r! * (n-r)!)
In the given expression (1 + 2x + 3x^2 + ...)^1/2, we have a = 1 and b = 2x + 3x^2 + ... Therefore, applying the binomial theorem, the expansion becomes:
(1 + 2x + 3x^2 + ...)^1/2 = C(1/2, 0) * 1^(1/2) * (2x + 3x^2 + ...)^0 + C(1/2, 1) * 1^(1/2 - 1) * (2x + 3x^2 + ...)^1/2 + ...
Simplifying this expression, we find that the coefficient of xn is given by:
C(1/2, n) * 1^(1/2 - n) * (2x + 3x^2 + ...)^n/2
Since C(1/2, n) = 1 for all values of n, and 1^(1/2 - n) = 1, the coefficient of xn simplifies to:
(2x + 3x^2 + ...)^n/2
From this, we can see that the coefficient of xn is simply the expression (2x + 3x^2 + ...)^n/2. Since this expression does not depend on n, the coefficient of xn is constant and equal to 1.
Therefore, the correct answer is option A) 1.
contains x4 then r is equal to
If rth ,(r+1)th and (r+2)th terms in the expansion of (1+x)n are in A.P. then- a)(n+2r)3 = n−2
- b)(n−2r)2 = n+2
- c)(n+2r)2 = n+2
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
If rth ,(r+1)th and (r+2)th terms in the expansion of (1+x)n are in A.P. then
a)
(n+2r)3 = n−2
b)
(n−2r)2 = n+2
c)
(n+2r)2 = n+2
d)
none of these
|
Mayank Dasgupta answered |
To determine the relationship between the terms (rth, (r+1)th, and (r+2)th) in the expansion of (1+x)^n, we can use the binomial theorem.
The general term of the expansion of (1+x)^n is given by:
C(n, r)*x^r*(1)^n-r
The rth term is given by:
C(n, r)*x^r*(1)^(n-r)
The (r+1)th term is given by:
C(n, r+1)*x^(r+1)*(1)^(n-r-1)
The (r+2)th term is given by:
C(n, r+2)*x^(r+2)*(1)^(n-r-2)
To determine if these terms are in an arithmetic progression (A.P.), we need to find the common difference between the terms. We can subtract consecutive terms to find the common difference:
(r+1)th term - rth term:
[C(n, r+1)*x^(r+1)*(1)^(n-r-1)] - [C(n, r)*x^r*(1)^(n-r)]
Simplifying this expression, we get:
C(n, r+1)*x^(r+1)*(1)^(n-r-1) - C(n, r)*x^r*(1)^(n-r)
= [C(n, r+1)*x^(r+1)*(1)^(n-r-1) - C(n, r)*x^r*(1)^(n-r)] / [x^r*(1)^(n-r)]
Similarly, the (r+2)th term - (r+1)th term can be found as:
[C(n, r+2)*x^(r+2)*(1)^(n-r-2)] - [C(n, r+1)*x^(r+1)*(1)^(n-r-1)]
Simplifying this expression, we get:
C(n, r+2)*x^(r+2)*(1)^(n-r-2) - C(n, r+1)*x^(r+1)*(1)^(n-r-1)
= [C(n, r+2)*x^(r+2)*(1)^(n-r-2) - C(n, r+1)*x^(r+1)*(1)^(n-r-1)] / [x^(r+1)*(1)^(n-r-1)]
Since we want the terms to be in an arithmetic progression, the common difference between these terms should be the same. Therefore, we can equate the expressions for the common difference and solve for x:
[C(n, r+1)*x^(r+1)*(1)^(n-r-1) - C(n, r)*x^r*(1)^(n-r)] / [x^r*(1)^(n-r)]
= [C(n, r+2)*x^(r+2)*(1)^(n-r-2) - C(n, r+1)*x^(r+1)*(1)^(n-r-1)] / [x^(r+1)*(1)^(n-r-1)]
Simplifying this expression and canceling out the common factors, we get:
[C(n, r+1)*x -
The general term of the expansion of (1+x)^n is given by:
C(n, r)*x^r*(1)^n-r
The rth term is given by:
C(n, r)*x^r*(1)^(n-r)
The (r+1)th term is given by:
C(n, r+1)*x^(r+1)*(1)^(n-r-1)
The (r+2)th term is given by:
C(n, r+2)*x^(r+2)*(1)^(n-r-2)
To determine if these terms are in an arithmetic progression (A.P.), we need to find the common difference between the terms. We can subtract consecutive terms to find the common difference:
(r+1)th term - rth term:
[C(n, r+1)*x^(r+1)*(1)^(n-r-1)] - [C(n, r)*x^r*(1)^(n-r)]
Simplifying this expression, we get:
C(n, r+1)*x^(r+1)*(1)^(n-r-1) - C(n, r)*x^r*(1)^(n-r)
= [C(n, r+1)*x^(r+1)*(1)^(n-r-1) - C(n, r)*x^r*(1)^(n-r)] / [x^r*(1)^(n-r)]
Similarly, the (r+2)th term - (r+1)th term can be found as:
[C(n, r+2)*x^(r+2)*(1)^(n-r-2)] - [C(n, r+1)*x^(r+1)*(1)^(n-r-1)]
Simplifying this expression, we get:
C(n, r+2)*x^(r+2)*(1)^(n-r-2) - C(n, r+1)*x^(r+1)*(1)^(n-r-1)
= [C(n, r+2)*x^(r+2)*(1)^(n-r-2) - C(n, r+1)*x^(r+1)*(1)^(n-r-1)] / [x^(r+1)*(1)^(n-r-1)]
Since we want the terms to be in an arithmetic progression, the common difference between these terms should be the same. Therefore, we can equate the expressions for the common difference and solve for x:
[C(n, r+1)*x^(r+1)*(1)^(n-r-1) - C(n, r)*x^r*(1)^(n-r)] / [x^r*(1)^(n-r)]
= [C(n, r+2)*x^(r+2)*(1)^(n-r-2) - C(n, r+1)*x^(r+1)*(1)^(n-r-1)] / [x^(r+1)*(1)^(n-r-1)]
Simplifying this expression and canceling out the common factors, we get:
[C(n, r+1)*x -
The coefficient of x99 in (x+1)(x+3)(x+5)………..(x+199) is- a)1.3.5…………..199
- b)1+3+5+……….+199
- c)1+2+3+………+99
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
The coefficient of x99 in (x+1)(x+3)(x+5)………..(x+199) is
a)
1.3.5…………..199
b)
1+3+5+……….+199
c)
1+2+3+………+99
d)
none of these
|
Avantika Joshi answered |
Method to Solve:
Max power of x is 100
So x^99 is nothing but sum of the roots of that equation
So which will be equal to -(1+3+5…+199)
+ ........, then z2 + 2z is equal to
is equal to 
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