Test: Binomial Theorem- 2 - JEE MCQ

# Test: Binomial Theorem- 2 - JEE MCQ

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## 25 Questions MCQ Test - Test: Binomial Theorem- 2

Test: Binomial Theorem- 2 for JEE 2024 is part of JEE preparation. The Test: Binomial Theorem- 2 questions and answers have been prepared according to the JEE exam syllabus.The Test: Binomial Theorem- 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Binomial Theorem- 2 below.
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Test: Binomial Theorem- 2 - Question 1

### The coefficient of second, third and fourth terms in the binomial expansion of (1+x)n(‘n’, a + ve integer) are in A.P.., if n is equal to

Detailed Solution for Test: Binomial Theorem- 2 - Question 1

/* We know that,
If a,b,c are in A.P then
a+c = 2b */

/* Splitting the middle term, we get

Here , n=2 is not possible because number of terms are more than three (According to the problem )
Therefore, n = 7

Test: Binomial Theorem- 2 - Question 2

### The coefficient of x3 in the binomial expansion of

Detailed Solution for Test: Binomial Theorem- 2 - Question 2

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

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Test: Binomial Theorem- 2 - Question 3

### The coefficient of x17 in the expansion of (x- 1) (x- 2) …..(x – 18) is

Detailed Solution for Test: Binomial Theorem- 2 - Question 3

The coefficient of x17 is given by
−1 + (−2) + (−3) + ….. (−18)
= −1 − 2 − 3….. − 18
= − (18(18+1))/2
= − 9(19)
= − 171

Test: Binomial Theorem- 2 - Question 4

If the coefficients of (r +1)th term and (r + 3)th term in the expansion of (1+x)2n be equal then

Detailed Solution for Test: Binomial Theorem- 2 - Question 4

Tr+1 = 2nCr(x)r (1)2n-r
Tr+3 = 2nCr+2 (x)r+2 (1)2n-r-2
Tr+1 : Tr+3
= 2nCr = 2nCr+2
=> 2n!/(2n-r)!r! = 2n!/(r+2)!(2n-r-2)!
=> 1/(2n-r)(2n-r-1) = 1/(r+2)(r+1)
=> (r+2)(r+1) = (2n-r)(2n-r-1)
=> r2 + 3r + 2 = 4n2 - 2nr - 2n - 2nr + r^2 + r
=> 0 = 4n2 - 4nr - 2n + r - 3r - 2
=> 0 = 4n2 - 4nr - 2n + r - 3r - 2 - 2 + 2
=> 0 = 4n2 - 4nr - 4 - [2n + 2r - 2]
=> 4n(n-r-1) -2(n-r-1)
Therefore n - r - 1 = 0
=> n = r+1

Test: Binomial Theorem- 2 - Question 5

If x = 9950+1005 and y = (101)50, then

Test: Binomial Theorem- 2 - Question 6

The greatest coefficient in the expansion of (1+x)12 is

Test: Binomial Theorem- 2 - Question 7

In the expansion of (1+x)60, the sum of coefficients of odd powers of x is

Detailed Solution for Test: Binomial Theorem- 2 - Question 7

(1+x)n= nC0+ nC1x+ nC2x2+...+ nC0xn ....(1)
(1−x)n= nC0 − nC1x+ nC2x2−...+ nC0 xn....(2)
⇒(1+x)n−(1−x)n = nC0+ nC1x+ nC2x2+...+ nC0xn− nC0+ nC1x− nC2x2+...+ nC0xn
=2(nC1x+nC3x3+....+xn)

Given: n=60 and put x=1
⇒(1+1)60−(1−1)60
=260
⇒260=2(nC1x+nC3x3+....+xn)

∴Sum of odd powers of x in the expansion is = nC1x+nC3x3+....+xn =260−1 =259

Test: Binomial Theorem- 2 - Question 8

Let n ∈ Q an n ∉ N,n ≠ 0, a > 0, then the expansion of (a+x)n in powers of x is valid if

Test: Binomial Theorem- 2 - Question 9

The expansion of , in powers of x, is valid if

Detailed Solution for Test: Binomial Theorem- 2 - Question 9

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

Test: Binomial Theorem- 2 - Question 10

The middle term in the expansion of (1+x)2n is

Detailed Solution for Test: Binomial Theorem- 2 - Question 10

Middle term in the expansion of (1+x)2n; is
=tn+1 = 2nCn.1(2n−n).xn
= {(2n)!.xn}/(2n−n)!n!
= {{2n(2n−1)(2n−2)(2n−3)....4×3×2×1}/n! n!}/xn
= {{2n[n(n−1)(n−2)....×2×1][(2n−1)(2n−3)....3×1]}/n! n!}xn
= [[(2n−1)(2n−3)....3×1]/n!] 2nxn

Test: Binomial Theorem- 2 - Question 11

The coefficient of x99 in (x+1)(x+3)(x+5)………..(x+199) is

Detailed Solution for Test: Binomial Theorem- 2 - Question 11

(x+1)(x+3)(x+5)(x+7)...(x+199)
We have (x+1)(x+3)=x2+(1+3)x+3
(x+1)(x+3)(x+5)=x^3+(1+3+5)x2 +(3+15+5)x+15
here Maximum power of x can be 100
For x^99 we have to multiply x from 99 terms and constant from remaning 1 term
So, when constant will be taken from (x+1) coff. of \$\$x99 will be 1
when constant term will be taken from (x+3) coff. of x99 will be 3
Similarly, When constant term will be taken from (x+199) coff. of x99 will be 199.
∴ coefficient of x^99will be sum of all these coffecient=1+3+5+..+199

Test: Binomial Theorem- 2 - Question 12

If the expansion of in powers of x contains the term x2r, then n−2r is

Detailed Solution for Test: Binomial Theorem- 2 - Question 12

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)

Test: Binomial Theorem- 2 - Question 13

Detailed Solution for Test: Binomial Theorem- 2 - Question 13

Test: Binomial Theorem- 2 - Question 14

If rth ,(r+1)th and (r+2)th terms in the expansion of (1+x)n are in A.P. then

Test: Binomial Theorem- 2 - Question 15

Coefficient of a2b5 in the expansion of (a+b)3(a−2b)4 is

Detailed Solution for Test: Binomial Theorem- 2 - Question 15

Test: Binomial Theorem- 2 - Question 16

In the expansion of(1+x)11, the 5th term is 24 times the 3rd term . The value of x is

Test: Binomial Theorem- 2 - Question 17

The sum of coefficients in the expansion of (x+2y+z)n is (n being a positive integer)

Test: Binomial Theorem- 2 - Question 18

If the rth term in the expansion of  contains x4 then r is equal to

Test: Binomial Theorem- 2 - Question 19

If a + b = 1, then  is equal to

Test: Binomial Theorem- 2 - Question 20

The 1st three terms in the expansion of (4 + x)3/2 are

Test: Binomial Theorem- 2 - Question 21

The coefficients of xn in the expansion of (1+2x + 3x2 + ........)1/2 is

Test: Binomial Theorem- 2 - Question 22

If z =  + ........, then z2 + 2z is equal to

Test: Binomial Theorem- 2 - Question 23

The index of the power of x that occurs in the 7th term from the end in the expansion of

Test: Binomial Theorem- 2 - Question 24

The index of the power of x that occurs in the 6th term in the expansion of

Detailed Solution for Test: Binomial Theorem- 2 - Question 24

Test: Binomial Theorem- 2 - Question 25

If in the expansion of(1+x)43, the coefficients of (2r+1)th and (r+2)th terms are equal, then r is equal to

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