Practice question of Binomial Theorem for iitjee , nda and airforce (Set 3)

Binomial Theorem

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Question no. 31

The first 3 terms in the expansion of (1+ax)ⁿ are 1 , 6x , 16x². then the value of a and n are respectively

looks_one 2 and 9

looks_two 3 and 2

looks_3 2/3 and 9

looks_4 3/2 and 6

Option looks_3 2/3 and 9

Solution :
First term = ⁿC₀(1)ⁿ(ax)⁰ = 1
Second term = ⁿC₁(1)^n-1(ax)¹ = nax
nax = 6x ⇒ na = 6 ⇒ a = 6/n
Third term = ⁿC₂(1)^n-2(ax)²

Question no. 32

Coefficients of x in the expansion of

looks_one 9a²

looks_two 10a³

looks_3 10a²

looks_4 10a

Option looks_two 10a³

Solution :

Question no. 33

In the expansion of , the coefficient of x^-10 will be

looks_one 12a¹¹

looks_two12b¹¹a

looks_3 12a¹¹b

looks_4 12a¹¹b¹¹

Option looks_3 12a¹¹b

Solution :

Question no. 34

The ratio of the coefficient of terms x^n-ra^r and x^ra^n-r in the binomial expansion of (x + a )ⁿ will be is

looks_one x : a

looks_two n : r

looks_3 x : n

looks_4 None of these

Option looks_4 None of these

Solution :
coefficient of terms x^n-ra^r = ⁿC_r = n!/(n-r)!(r!)
coefficient of terms x^ra^n-r = ⁿC_n-r = = n!/(n-r)!(r!)
Both are same

Question no. 35

If the expansion of , the coefficient of y will be

looks_one 20c

looks_two 10c

looks_3 10c³

looks_4 20c²

Option looks_3 10c³

Solution :

Question no. 36

In the expansion of , the constant term is

looks_one -20

looks_two 20

looks_3 30

looks_4 -30

Option looks_one -20

Solution :

Question no. 37

In the expansion of , the coefficient of x⁴ is

looks_one

looks_two

looks_3

looks_4 None of these

Option looks_one

Solution :

Question no. 38

If the coefficients of 5^th , 6^th and 7^th terms in the expansion of (1+x)ⁿ be in A.P. , then n =

looks_one7 only

looks_two 14 only

looks_37 or 14

looks_4 None of these

Option looks_37 or 14

Solution :
T₅ = ⁿC₄(x)⁴
T₆ = ⁿC₅(x)⁵
T₇ = ⁿC₆(x)⁶
According to Question
T₅ + T₇ = 2T₆
ⁿC₄ + ⁿC₆ = 2ⁿC₅

Question no. 39

The coefficient of x^-7 in the expansion of will be

looks_one

looks_two

looks_3

looks_4

Option looks_two

Solution :

Question no. 40

The coefficient of x⁵³ in the following expansion is

looks_one¹⁰⁰C₄₇

looks_two¹⁰⁰C₅₃

looks_3- ¹⁰⁰C₅₃

looks_4- ¹⁰⁰C₁₀₀

Option looks_3- ¹⁰⁰C₅₃

Solution :

It can be written in following format :

Question no. 41

If the coefficient of x⁷ and x⁸ in are equal , then n is

looks_one 56

looks_two 55

looks_3 45

looks_4 15

Option looks_two 55

Solution :

Question no. 42

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

looks_one 6

looks_two 9

looks_3 12

looks_4 24

Option looks_3 12

Solution :
( 1+ x )^m ( 1 - x )ⁿ = (^mC₀ + ^mC₁x + ^mC₂x² + ----- )(ⁿC₀ - ⁿC₁x + ⁿC₂x² + ----- )
Coefficient of x = - ^mC₀ⁿC₁ + ^mC₁ⁿC₀ = -n + m = 3
Coefficient of x² = ⁿC₂ - ^mC₁ⁿC₁ + ^mC₂ = -6

Question no. 43

If the coefficient of 2^nd , 3 ^rd and 4^thterms in the binomial expansion of (1+x)ⁿ are in A.P. then n² -9n is equal to

looks_one -7

looks_two 7

looks_3 14

looks_4 -14

Option looks_4 -14

Solution :
T₂ = ⁿC₁(1)^n-1(x)¹
T₃ = ⁿC₂(1)^n-2(x)²
T₄ = ⁿC₃(1)^n-3(x)³
Coefficient of T₂ = ⁿC₁ = n
Coefficient of T₃ = ⁿC₂ = n(n-1)/2
Coefficient of T₄ = ⁿC₃ = n(n-1)(n-2)/6
Coefficient are in AP
2*n(n-1)/2 = n + n(n-1)(n-2)/6
n(n-1) = n + n(n-1)(n-2)/6
n-1 = 1 + (n-1)(n-2)/6
6n -6 = 6 + n² + 2 -3n
-14 = n² - 9n

Question no. 44

If the coefficient of 4^th term in the expansion of (a+b)ⁿ is 56, then n is

looks_one 12

looks_two 10

looks_3 8

looks_4 6

Option looks_3 8

Solution :

Question no. 45

The coefficient of x⁵ in the expansion of (1 + x)²¹ + (1 + x)²² + ....... +(1 + x)³⁰ is

looks_one ⁵¹C₅

looks_two ⁹C₅

looks_3³¹C₆ -²¹C₆

looks_4 ³⁰C₅ + ²⁰C₅

Option looks_3³¹C₆ -²¹C₆

Solution :
coefficient of x⁵ = ²¹C₅ + ²²C₅ + ²³C₅ + ....... +³⁰C₅
ⁿC_n + ⁿ⁺¹C_n + ⁿ⁺²C_n + ....... +²ⁿC_n = ²ⁿ⁺¹C_n+1
⁵C₅ + ⁶C₅ + ⁷C₅ + ....... +³⁰C₅ = ³¹C₆ .......[1]
⁵C₅ + ⁶C₅ + ⁷C₅ + ....... +²⁰C₅ = ²¹C₆ .......[2]
subtract [1] - [2]
²¹C₅ + ²²C₅ + ²³C₅ + ....... +³⁰C₅ = ³¹C₆ - ²¹C₆

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