Practice Start Test

Binomial Theorem

JEE Mathematics · 151 questions · Page 13 of 16 · Click an option or "Show Solution" to reveal answer

Q121
The positive integer just greater than (1+0.0001)10000{\left( {1 + 0.0001} \right)^{10000}} is
A 4
B 5
C 2
D 3
Correct Answer
Option D
Solution
(1+0.0001)10000{\left( {1 + 0.0001} \right)^{10000}}

=

(1+1104)10000{\left( {1 + {1 \over {{{10}^4}}}} \right)^{10000}}

= 1 + 10000

×1104{ \times {1 \over {{{10}^4}}}}

+

10000(9999)2!×(1104)2{{10000\left( {9999} \right)} \over {2!}} \times {\left( {{1 \over {{{10}^4}}}} \right)^2}

+......∞\infty < 1 + 1 +

12!{1 \over {2!}}

+

13!{1 \over {3!}}

+ ...... ∞\infty = e = 2.71828 < 3

Q122
If some three consecutive in the binomial expansion of (x + 1)n is powers of x are in the ratio 2 : 15 : 70, then the average of these three coefficient is :-
A 625
B 227
C 964
D 232
Correct Answer
Option D
Solution

Given

nCr−1:nCr:nCr+1=2:15:70{}^n{C_{r - 1}}:{}^n{C_r}:{}^n{C_{r + 1}} = 2:15:70
nCr−1nCr=215{{{}^n{C_{r - 1}}} \over {{}^n{C_r}}} = {2 \over {15}}
⇒rn−r+1=215\Rightarrow {r \over {n - r + 1}} = {2 \over {15}}
⇒15r=2n−2r+2\Rightarrow 15r = 2n - 2r + 2
⇒17r=2n+2\Rightarrow 17r = 2n + 2

.... (i) Now

nCrnCr+1=1570{{{}^n{C_r}} \over {{}^n{C_{r + 1}}}} = {{15} \over {70}}
⇒r+1n−r=314\Rightarrow {{r + 1} \over {n - r}} = {3 \over {14}}

⇒\Rightarrow 14r + 14 = 3n – 3r ⇒\Rightarrow 17r = 3n – 14 ... (ii) Now From (i) and (ii) equation 2n + 2 = 3n - 14 ⇒\Rightarrow n = 16 By putting n = 16 in equation (i) ⇒\Rightarrow r = 2 ∴\therefore Average of coefficient =

16C1+16C2+16C33{{{}^{16}{C_1} + {}^{16}{C_2} + {}^{16}{C_3}} \over 3}

=

16+120+5603{{16 + 120 + 560} \over 3}

=

6963=232{{696} \over 3} = 232
Q123
If Sn=∑r=0n1nCr  and  tn=∑r=0nrnCr, {S_n} = \sum\limits_{r = 0}^n {{1 \over {{}^n{C_r}}}} \,\,and\,\,{t_n} = \sum\limits_{r = 0}^n {{r \over {{}^n{C_r}}},\,} then tnSn{{{t_{ n}}} \over {{S_n}}} is equal to
A 2n−12{{2n - 1} \over 2}
B 12n−1{1 \over 2}n - 1
C n - 1
D 12n{1 \over 2}n
Correct Answer
Option D
Solution
Sn=∑r=0n1nCr{S_n} = \sum\limits_{r = 0}^n {{1 \over {{}^n{C_r}}}}

=

1nC0+1nC1+....+1nCn{1 \over {{}^n{C_0}}} + {1 \over {{}^n{C_1}}} + .... + {1 \over {{}^n{C_n}}}
tn=∑r=0nrnCr{t_n} = \sum\limits_{r = 0}^n {{r \over {{}^n{C_r}}}}

=

0nC0+1nC1+2nC2....+nnCn{0 \over {{}^n{C_0}}} + {1 \over {{}^n{C_1}}} + {2 \over {{}^n{C_2}}}.... + {n \over {{}^n{C_n}}}

.........(1) We can write

tn{t_n}

by rearranging like this,

tn{t_n}

=

nnCn+n−1nCn−1+...+1nC1+0nC0{n \over {{}^n{C_n}}} + {{n - 1} \over {{}^n{C_{n - 1}}}} + ... + {1 \over {{}^n{C_1}}} + {0 \over {{}^n{C_0}}}

=

nnC0+n−1nC1+...+1nCn−1+0nCn{n \over {{}^n{C_0}}} + {{n - 1} \over {{}^n{C_1}}} + ... + {1 \over {{}^n{C_{n - 1}}}} + {0 \over {{}^n{C_n}}}

.........(2) [as

nC0{{}^n{C_0}}

=

nCn{{}^n{C_n}}

,

nC1{{}^n{C_1}}

=

nCn−1{{}^n{C_{n - 1}}}

......] By adding (1) and (2) we get,

2tn2{t_n}

=

nnC0+nnC1+...+nnCn−1+nnCn{n \over {{}^n{C_0}}} + {n \over {{}^n{C_1}}} + ... + {n \over {{}^n{C_{n - 1}}}} + {n \over {{}^n{C_n}}}

=

n[1nC0+1nC1+...+1nCn−1+1nCn]n\left[ {{1 \over {{}^n{C_0}}} + {1 \over {{}^n{C_1}}} + ... + {1 \over {{}^n{C_{n - 1}}}} + {1 \over {{}^n{C_n}}}} \right]

= n

Sn{S_n}

∴\therefore

tnSn=n2{{{t_n}} \over {{S_n}}} = {n \over 2}
Q124
If the coefficients of rth, (r+1)th, and (r + 2)th terms in the binomial expansion of (1+y)m{{\rm{(1 + y )}}^m} are in A.P., then m and r satisfy the equation
A m2−m(4r−1)+4 r2−2=0{m^2} - m(4r - 1) + 4\,{r^2} - 2 = 0
B m2−m(4r+1)+4 r2+2=0{m^2} - m(4r + 1) + 4\,{r^2} + 2 = 0
C m2−m(4r+1)+4 r2−2=0{m^2} - m(4r + 1) + 4\,{r^2} - 2 = 0
D m2−m(4r−1)+4 r2+2=0{m^2} - m(4r - 1) + 4\,{r^2} + 2 = 0
Correct Answer
Option C
Solution

Let r = 2 ∴\therefore 2nd, 3rd and 4th terms are in AP. 2nd term = T2 =

mC1.y{}^m{C_1}.y

Coefficient of T2 =

mC1{}^m{C_1}

3rd term = T3 =

mC2.y2{}^m{C_2}.{y^2}

Coefficient of T3 =

mC2{}^m{C_2}

4th term = T4 =

mC3.y3{}^m{C_3}.{y^3}

Coefficient of T2 =

mC3{}^m{C_3}

∴\therefore 2.

mC2{}^m{C_2}

=

mC1{}^m{C_1}

+

mC3{}^m{C_3}

⇒\Rightarrow

2.m(m−1)1.22.{{m\left( {m - 1} \right)} \over {1.2}}

=

m1{m \over 1}

+

m(m−1)(m−2)1.2.3{{m\left( {m - 1} \right)\left( {m - 2} \right)} \over {1.2.3}}

⇒\Rightarrow 6m2 - 6m = 6m +m(m2 - 3m + 2) ⇒\Rightarrow 6m2 - 6m = 6m + m3 - 3m2 + 2m ⇒\Rightarrow 6m - 6 = 6 + m2 - 3m + 2 ⇒\Rightarrow m2 - 9m + 14 = 0 Now put r = 2 at each option and find answer.

In option C,

m2−m(4r+1)+4 r2−2=0{m^2} - m(4r + 1) + 4\,{r^2} - 2 = 0

putting r = 2 we get m2 - 9m + 14 = 0. So Option C is correct.

Q125
The sum of the series 2.20C0 + 5.20C1 + 8.20C2 + 11.20C3 + ... +62.20C20 is equal to :
A 225
B 224
C 226
D 223
Correct Answer
Option A
Solution

Here general term = (3r + 2)20Cr ∴\therefore Sum of the series =

∑r=020(3r+2)20Cr\sum\limits_{r = 0}^{20} {\left( {3r + 2} \right)} {}^{20}{C_r}

=

3∑r=020r.20Cr+2∑r=02020Cr3\sum\limits_{r = 0}^{20} {r.} {}^{20}{C_r} + 2\sum\limits_{r = 0}^{20} {{}^{20}{C_r}}

= 3 ×\times 20×\times220 - 1 + 2×\times220 = 60×\times219 + 221 = 221 [15 + 1] = 221 ×\times 16 = 225

Q126
If the sum of the coefficients in the expansion of  (a+b)n\,{\left( {a + b} \right)^n} is 4096, then the greatest coefficient in the expansion is
A 1594
B 792
C 924
D 2924
Correct Answer
Option C
Solution

We know,

 (a+b)n\,{\left( {a + b} \right)^n}

=

nC0.an+nC1.an−1.b+...+nCn.bn{}^n{C_0}.{a^n} + {}^n{C_1}.{a^{n - 1}}.b + ... + {}^n{C_n}.{b^n}

Remember to find sum of coefficient of binomial expansion we ave to put 1 in place of all the variable. So put

aa

= b = 1 ∴\therefore 2n =

nC0+nC1+nC2...+nCn{}^n{C_0} + {}^n{C_1} + {}^n{C_2}... + {}^n{C_n}

According to question, 2n = 4096 = 212

⇒n=12\Rightarrow n = 12

So

 (a+b)n\,{\left( {a + b} \right)^n}

=

 (a+b)12\,{\left( {a + b} \right)^{12}}

Here n = 12 is even so formula for greatest term is

Tn2+1=nCn2.an2.bn2{T_{{n \over 2} + 1}} = {}^n{C_{{n \over 2}}}.{a^{{n \over 2}}}.{b^{{n \over 2}}}

For n = 12, greatest term

T6+1=12C6.a6.b6{T_{6 + 1}} = {}^{12}{C_6}.{a^6}.{b^6}

∴\therefore Coefficient of the greatest term =

12C6{}^{12}{C_6}

=

12!6!6!{{12!} \over {6!6!}}

= 924

Q127
For natural numbers mm , nn, if (1−y)m(1+y)n  =1+a1y+a2y2+..........{\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n}\,\, = 1 + {a_1}y + {a_2}{y^2} + .......... and a1=a2=10,{a_1} = {a_2} = 10, then (m, n)\left( {m,\,n} \right) is
A (20, 45)\left( {20,\,45} \right)
B (35, 20)\left( {35,\,20} \right)
C (45, 35)\left( {45,\,35} \right)
D (35, 45)\left( {35,\,45} \right)
Correct Answer
Option D
Solution
(1−y)m(1+y)n  {\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n}\,\,

=

(mC0−mC1y+mC2y2+....)\left( {{}^m{C_0} - {}^m{C_1}y + {}^m{C_2}{y^2} + ....} \right)

-

(nC0+nC1y+nC2y2+....)\left( {{}^n{C_0} + {}^n{C_1}y + {}^n{C_2}{y^2} + ....} \right)
a1{a_1}

= Coefficient of y =

nC1{{}^n{C_1}}

-

mC1{{}^m{C_1}}

= 10 ⇒\Rightarrow n - m = 10

a2{a_2}

= Coefficient of y2 =

nC2+nC1×mC1+mC2=10{}^n{C_2} + {}^n{C_1} \times {}^m{C_1} + {}^m{C_2} = 10
⇒n(n−1)2−nm+m(m−1)2=10\Rightarrow {{n\left( {n - 1} \right)} \over 2} - nm + {{m\left( {m - 1} \right)} \over 2} = 10
⇒n(n−1)−2nm+m(m−1)=20\Rightarrow n\left( {n - 1} \right) - 2nm + m\left( {m - 1} \right) = 20

⇒\Rightarrow (m + 10)(m + 9) - 2(m + 10)m + m(m - 1) = 20 ⇒\Rightarrow 90 + 19m + m2 - 2m2 - 20m + m2 - m - 20 = 0 ⇒\Rightarrow 70 - 2m = 0 ⇒\Rightarrow m = 35 ∴\therefore n = 10 + 35 = 45

Q128
For x ∈ \in R, x ≠ \ne -1, if (1 + x)2016 + x(1 + x)2015 + x2(1 + x)2014 + . . . . + x2016 = ∑i=02016ai xi,  \sum\limits_{i = 0}^{2016} {{a_i}} \,{x^i},\,\, then a17 is equal to :
A 2017!17!   2000!{{2017!} \over {17!\,\,\,2000!}}
B 2016!17!   1999!{{2016!} \over {17!\,\,\,1999!}}
C 2017!2000!{{2017!} \over {2000!}}
D 2016!16!{{2016!} \over {16!}}
Correct Answer
Option A
Solution

Assume, P = (1 + x)2016 + x(1 + x)2015 + . . . . .+ x2015 . (1 + x) + x2016 . . . . .(1) Multiply this with

(x1+x),\left( {{x \over {1 + x}}} \right),
(x1+x)P=\left( {{x \over {1 + x}}} \right)P =

x(1 + x)2015 + x2(1 + x)2014 + . . . . . . + x2016 +

x20171+x{{{x^{2017}}} \over {1 + x}}

. . . . . (2) Performing (1) −- (2), we get

P1+x={P \over {1 + x}} =

(1 + x)2016 −-

x20171+x{{{x^{2017}}} \over {1 + x}}

⇒\Rightarrow P = (1 + x)2017 −- x2017 ∴\therefore a17 = coefficient of x17 == 2017C17 ==

2017!17!  2000!{{2017!} \over {17!\,\,2000!}}
Q129
If the constant term in the expansion of (3x3−2x2+5x5)10{\left( {3{x^3} - 2{x^2} + {5 \over {{x^5}}}} \right)^{10}} is 2k.l, where l is an odd integer, then the value of k is equal to:
A 6
B 7
C 8
D 9
Correct Answer
Option D
Solution

Note : Multinomial Theorem : The general term of

(x1+x2+...+xn)n{\left( {{x_1} + {x_2} + ... + {x_n}} \right)^n}

the expansion is

n!n1!n2!...nn!x1n1x2n2...xnnn{{n!} \over {{n_1}!{n_2}!...{n_n}!}}x_1^{{n_1}}x_2^{{n_2}}...x_n^{{n_n}}

where n1 + n2 + ..... + nn = n Given,

(3x2−2x2+5x5)10{\left( {3{x^2} - 2{x^2} + {5 \over {{x^5}}}} \right)^{10}}
=(3x8−2x7+5)10x50= {{{{(3{x^8} - 2{x^7} + 5)}^{10}}} \over {{x^{50}}}}

Now constant term in

(3x3−2x2+5x5)10=x50{\left( {3{x^3} - 2{x^2} + {5 \over {{x^5}}}} \right)^{10}} = {x^{50}}

term in

(3x8−2x7+5)10{(3{x^8} - 2{x^7} + 5)^{10}}

General term in

(3x8−2x7+5)10{(3{x^8} - 2{x^7} + 5)^{10}}

is

=10!n1! n2! n3!(3x8)n1(−2x7)n2(5)n3= {{10!} \over {{n_1}!\,{n_2}!\,{n_3}!}}{(3{x^8})^{{n_1}}}{( - 2{x^7})^{{n_2}}}{(5)^{{n_3}}}
=10!n1! n2! n3!(3)n1(−2)n2(5)n3 . x8n1+7n2= {{10!} \over {{n_1}!\,{n_2}!\,{n_3}!}}{(3)^{{n^1}}}{( - 2)^{{n_2}}}{(5)^{{n^3}}}\,.\,{x^{8{n_1} + 7{n_2}}}

∴\therefore Coefficient of

x8n1+7n2{x^{8{n_1} + 7{n_2}}}

is

=10!n1! n2! n3!(3)n1(−2)n2(5)n3= {{10!} \over {{n_1}!\,{n_2}!\,{n_3}!}}{(3)^{{n_1}}}{( - 2)^{{n_2}}}{(5)^{{n_3}}}

where

n1+n2+n3=0{n_1} + {n_2} + {n_3} = 0

For coefficient of x50 :

8n1+7n2=508{n_1} + 7{n_2} = 50

∴\therefore Possible values of n1, n2 and n3 are .tg .tg n

1_1

n

2_2

n

3_3

1 6 3 ∴\therefore Coefficient of x50

=10!1! 6! 3!(3)1(−2)6(5)3= {{10!} \over {1!\,6!\,3!}}{(3)^1}{( - 2)^6}{(5)^3}
=10×9×8×76×3×53×26= {{10 \times 9 \times 8 \times 7} \over 6} \times 3 \times {5^3} \times {2^6}
=5×3×8×7×3×53×26= 5 \times 3 \times 8 \times 7 \times 3 \times {5^3} \times {2^6}
=7×54×32×29= 7 \times {5^4} \times {3^2} \times {2^9}
=2k . l= {2^k}\,.\,l

∴\therefore

l=7×54×32l = 7 \times {5^4} \times {3^2}

= An odd integer and

2k=29{2^k} = {2^9}
⇒k=9\Rightarrow k = 9
Q130
If n is the number of irrational terms in the expansion of (31/4+51/8)60{\left( {{3^{1/4}} + {5^{1/8}}} \right)^{60}}, then (n −- 1) is divisible by :
A 30
B 8
C 7
D 26
Correct Answer
Option D
Solution
Tr+1=60Cr(31/4)60−r(51/8)r{T_{r + 1}} = {}^{60}{C_r}{\left( {{3^{1/4}}} \right)^{60 - r}}{\left( {{5^{1/8}}} \right)^r}

rational if

60−r4,r8{{60 - r} \over 4},{r \over 8}

, both are whole numbers,

r∈{0,1,2,......60}r \in \{ 0,1,2,......60\}
60−r4∈W⇒r∈{0,4,8,....60}{{60 - r} \over 4} \in W \Rightarrow r \in \{ 0,4,8,....60\}

and

r8∈W⇒r∈{0,8,16,.....56}{r \over 8} \in W \Rightarrow r \in \{ 0,8,16,.....56\}

∴\therefore Common terms

r∈{0,8,16,.....56}r \in \{ 0,8,16,.....56\}

So, 8 terms are rational Then Irrational terms =

61−8=53=n61 - 8 = 53 = n

∴\therefore

n−1=52=13×22n - 1 = 52 = 13 \times {2^2}

Factors 1, 2, 4, 13, 26, 52

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