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Binomial Theorem

JEE Mathematics · 151 questions · Page 5 of 16 · Click an option or "Show Solution" to reveal answer

Q41
A possible value of 'x', for which the ninth term in the expansion of {3log⁡325x−1+7+3(−18)log⁡3(5x−1+1)}10{\left\{ {{3^{{{\log }_3}\sqrt {{{25}^{x - 1}} + 7} }} + {3^{\left( { - {1 \over 8}} \right){{\log }_3}({5^{x - 1}} + 1)}}} \right\}^{10}} in the increasing powers of 3(−18)log⁡3(5x−1+1){3^{\left( { - {1 \over 8}} \right){{\log }_3}({5^{x - 1}} + 1)}} is equal to 180, is :
A 0
B −-1
C 2
D 1
Correct Answer
Option D
Solution
10C8(25(x−1)+7)×(5(x−1)+1)−1=180{}^{10}{C_8}({25^{(x - 1)}} + 7) \times {({5^{(x - 1)}} + 1)^{ - 1}} = 180
⇒25x−1+75(x−1)+1=4\Rightarrow {{{{25}^{x - 1}} + 7} \over {{5^{(x - 1)}} + 1}} = 4
⇒t2+7t+1=4\Rightarrow {{{t^2} + 7} \over {t + 1}} = 4

; ⇒\Rightarrow t = 1, 3 = 5x −- 1 ⇒\Rightarrow x −- 1 = 0 (one of the possible value). ⇒\Rightarrow x = 1

Q42
The sum of all those terms which are rational numbers in the expansion of (21/3 + 31/4)12 is :
A 89
B 27
C 35
D 43
Correct Answer
Option D
Solution
Tr+1=12Cr(21/3)r.(31/4)12−4{T_{r + 1}} = {}^{12}{C_r}{\left( {{2^{1/3}}} \right)^r}.{\left( {{3^{1/4}}} \right)^{12 - 4}}

Tr + 1 will be rational number when r = 0, 3, 6, 9, 12 & r = 0, 4, 8, 12 ⇒\Rightarrow r = 0, 12 T1 + T13 = 1 ×\times 33 + 1 ×\times 24 ×\times 1 = 24 + 16 = 43

Q43
If the greatest value of the term independent of 'x' in the expansion of (xsin⁡α+acos⁡αx)10{\left( {x\sin \alpha + a{{\cos \alpha } \over x}} \right)^{10}} is 10!(5!)2{{10!} \over {{{(5!)}^2}}}, then the value of 'a' is equal to :
A −-1
B 1
C −-2
D 2
Correct Answer
Option D
Solution
Tr+1=10Cr(xsin⁡α)10−r(acos⁡αx)r{T_{r + 1}} = {}^{10}{C_r}{(x\sin \alpha )^{10 - r}}{\left( {{{a\cos \alpha } \over x}} \right)^r}

r = 0, 1, 2, ......., 10 Tr + 1 will be independent of x when 10 −- 2r = 0 ⇒\Rightarrow r = 5

T6=10C5(xsin⁡α)5×(acos⁡αx)5{T_6} = {}^{10}{C_5}{(x\sin \alpha )^5} \times {\left( {{{a\cos \alpha } \over x}} \right)^5}
=10C5×a5×125(sin⁡2α)5= {}^{10}{C_5} \times {a^5} \times {1 \over {{2^5}}}{(\sin 2\alpha )^5}

will be greatest when sin2α\alpha = 1

⇒10C5a525=10C5⇒a=2\Rightarrow {}^{10}{C_5}{{{a^5}} \over {{2^5}}} = {}^{10}{C_5} \Rightarrow a = 2
Q44
The lowest integer which is greater than (1+110100)10100{\left( {1 + {1 \over {{{10}^{100}}}}} \right)^{{{10}^{100}}}} is ______________.
A 3
B 4
C 2
D 1
Correct Answer
Option A
Solution

Let

P=(1+110100)10100P = {\left( {1 + {1 \over {{{10}^{100}}}}} \right)^{{{10}^{100}}}}

Let

x=10100x = {10^{100}}
⇒P=(1+1x)x\Rightarrow P = {\left( {1 + {1 \over x}} \right)^x}
⇒P=1+(x)(1x)+(x)(x−1)∣ ⁣ 2 ‾.1x2+(x)(x−1)(x−2)∣ ⁣ 3 ‾.1x3+....\Rightarrow P = 1 + (x)\left( {{1 \over x}} \right) + {{(x)(x - 1)} \over {\left| \!{\underline {\, 2 \,}} \right. }}.{1 \over {{x^2}}} + {{(x)(x - 1)(x - 2)} \over {\left| \!{\underline {\, 3 \,}} \right. }}.{1 \over {{x^3}}} + ....

(upto 10100 + 1 terms)

⇒P=1+1+(1∣ ⁣ 2 ‾−1∣ ⁣ 2 ‾x2)+(1∣ ⁣ 3 ‾−...)+...\Rightarrow P = 1 + 1 + \left( {{1 \over {\left| \!{\underline {\, 2 \,}} \right. }} - {1 \over {\left| \!{\underline {\, 2 \,}} \right. {x^2}}}} \right) + \left( {{1 \over {\left| \!{\underline {\, 3 \,}} \right. }} - ...} \right) + ...

so on

⇒P=2+(Positive value less than 1∣ ⁣ 2 ‾+1∣ ⁣ 3 ‾+1∣ ⁣ 4 ‾+...)\Rightarrow P = 2 + \left( {Positive\,value\,less\,than\,{1 \over {\left| \!{\underline {\, 2 \,}} \right. }} + {1 \over {\left| \!{\underline {\, 3 \,}} \right. }} + {1 \over {\left| \!{\underline {\, 4 \,}} \right. }} + ...} \right)

Also

e=1+1∣ ⁣ 1 ‾+1∣ ⁣ 2 ‾+1∣ ⁣ 3 ‾+1∣ ⁣ 4 ‾+...e = 1 + {1 \over {\left| \!{\underline {\, 1 \,}} \right. }} + {1 \over {\left| \!{\underline {\, 2 \,}} \right. }} + {1 \over {\left| \!{\underline {\, 3 \,}} \right. }} + {1 \over {\left| \!{\underline {\, 4 \,}} \right. }} + ...
⇒1∣ ⁣ 2 ‾+1∣ ⁣ 3 ‾+1∣ ⁣ 4 ‾+...=e−2\Rightarrow {1 \over {\left| \!{\underline {\, 2 \,}} \right. }} + {1 \over {\left| \!{\underline {\, 3 \,}} \right. }} + {1 \over {\left| \!{\underline {\, 4 \,}} \right. }} + ... = e - 2

⇒\Rightarrow P = 2 + (positive value less than e −- 2) ⇒\Rightarrow P

∈\in

(2, 3) ⇒\Rightarrow least integer value of P is 3

Q45
∑k=020(20Ck)2\sum\limits_{k = 0}^{20} {{{\left( {{}^{20}{C_k}} \right)}^2}} is equal to :
A 40C21{}^{40}{C_{21}}
B 40C19{}^{40}{C_{19}}
C 40C20{}^{40}{C_{20}}
D 41C20{}^{41}{C_{20}}
Correct Answer
Option C
Solution
∑k=020(20Ck)2\sum\limits_{k = 0}^{20} {{{\left( {{}^{20}{C_k}} \right)}^2}}

=

(20C0)2+(20C1)2+(20C2)2+....+(20C20)2{\left( {{}^{20}{C_0}} \right)^2} + {\left( {{}^{20}{C_1}} \right)^2} + {\left( {{}^{20}{C_2}} \right)^2} + .... + {\left( {{}^{20}{C_{20}}} \right)^2}

= 40C20 Using the formula :

(nC0)2+(nC1)2+(nC2)2+....+(nCn)2=2nCn{\left( {{}^n{C_0}} \right)^2} + {\left( {{}^n{C_1}} \right)^2} + {\left( {{}^n{C_2}} \right)^2} + .... + {\left( {{}^n{C_n}} \right)^2} = {}^{2n}{C_n}
Q46
Let n ≥\ge 5 be an integer. If 9n −- 8n −- 1 = 64α\alpha and 6n −- 5n −- 1 = 25β\beta, then α\alpha −- β\beta is equal to
A 1 + nC2 (8 −- 5) + nC3 (82 −- 52) + ...... + nCn (8n −- 1 −- 5n −- 1)
B 1 + nC3 (8 −- 5) + nC4 (82 −- 52) + ...... + nCn (8n −- 2 −- 5n −- 2)
C nC3 (8 −- 5) + nC4 (82 −- 52) + ...... + nCn (8n −- 2 −- 5n −- 2)
D nC4 (8 −- 5) + nC5 (82 −- 52) + ...... + nCn (8n −- 3 −- 5n −- 3)
Correct Answer
Option C
Solution

Given,

9n−8n−1=64α{9^n} - 8n - 1 = 64\alpha
⇒α=(1+8)n−8n−164\Rightarrow \alpha = {{{{(1 + 8)}^n} - 8n - 1} \over {64}}
=(nC0 . 1+nC1 . 81+nC2 . 82  +  .....  +  nCn . 8n)−8n−182= {{\left( {{}^n{C_0}\,.\,1 + {}^n{C_1}\,.\,{8^1} + {}^n{C_2}\,.\,{8^2}\,\, + \,\,.....\,\, + \,\,{}^n{C_n}\,.\,{8^n}} \right) - 8n - 1} \over {{8^2}}}
=1+8n+nC2 . 82  +  ....  +  nCn . 8n−8n−182= {{1 + 8n + {}^n{C_2}\,.\,{8^2}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{8^n} - 8n - 1} \over {{8^2}}}
=nC2 . 82+nC3 . 83  +  .....  +  nCn . 8n82= {{{}^n{C_2}\,.\,{8^2} + {}^n{C_3}\,.\,{8^3}\,\, + \,\,.....\,\, + \,\,{}^n{C_n}\,.\,{8^n}} \over {{8^2}}}
=nC2+nC3 . 8+nC4 . 82  +  .....  nCn . 8n−2= {}^n{C_2} + {}^n{C_3}\,.\,8 + {}^n{C_4}\,.\,{8^2}\,\, + \,\,.....\,\,{}^n{C_n}\,.\,{8^{n - 2}}

Also given,

6n−5n−1=25β{6^n} - 5n - 1 = 25\beta
⇒β=(1+5)n−5n−125\Rightarrow \beta = {{{{(1 + 5)}^n} - 5n - 1} \over {25}}
=nC0 . 1+nC1 . 5+nC2 . 52  +  .....  +  nCn . 5n−5n−152= {{{}^n{C_0}\,.\,1 + {}^n{C_1}\,.\,5 + {}^n{C_2}\,.\,{5^2}\,\, + \,\,.....\,\, + \,\,{}^n{C_n}\,.\,{5^n} - 5n - 1} \over {{5^2}}}
=1+5n+nC2 . 52+nC3 . 53  +  ....  +  nCn . 52−5n−152= {{1 + 5n + {}^n{C_2}\,.\,{5^2} + {}^n{C_3}\,.\,{5^3}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{5^2} - 5n - 1} \over {{5^2}}}
=nC2 . 52+nC3 . 53+nC4 . 54  +  ....  +  nCn . 5n52= {{{}^n{C_2}\,.\,{5^2} + {}^n{C_3}\,.\,{5^3} + {}^n{C_4}\,.\,{5^4}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{5^n}} \over {{5^2}}}
=nC2+nC3 . 5+nC4 . 52  +  .....  +  nCn . 5n−2= {}^n{C_2} + {}^n{C_3}\,.\,5 + {}^n{C_4}\,.\,{5^2}\,\, + \,\,.....\,\, + \,\,{}^n{C_n}\,.\,{5^{n - 2}}

∴\therefore

α−β\alpha - \beta
=(nC2+nC3 . 8+nC4 . 82  +  ....  +  nCn . 8n−2)−(nC2+nC3 . 5+nC4 . 52  +  ....  +  nCn . 5n−2)= \left( {{}^n{C_2} + {}^n{C_3}\,.\,8 + {}^n{C_4}\,.\,{8^2}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{8^{n - 2}}} \right) - \left( {{}^n{C_2} + {}^n{C_3}\,.\,5 + {}^n{C_4}\,.\,{5^2}\,\, + \,\,....\,\, + \,\,{}^n{C_n}\,.\,{5^{n - 2}}} \right)
=nC3 . (8−5)+nC4 . (82−52)  +  ....  +  nCn(8n−2−5n−2)= {}^n{C_3}\,.\,(8 - 5) + {}^n{C_4}\,.\,({8^2} - {5^2})\,\, + \,\,....\,\, + \,\,{}^n{C_n}({8^{n - 2}} - {5^{n - 2}})
Q47
If ∑k=131(31Ck)(31Ck−1)−∑k=130(30Ck)(30Ck−1)=α(60!)(30!)(31!)\sum\limits_{k = 1}^{31} {\left( {{}^{31}{C_k}} \right)\left( {{}^{31}{C_{k - 1}}} \right) - \sum\limits_{k = 1}^{30} {\left( {{}^{30}{C_k}} \right)\left( {{}^{30}{C_{k - 1}}} \right) = {{\alpha (60!)} \over {(30!)(31!)}}} } , where α\alpha ∈\in R, then the value of 16α\alpha is equal to
A 1411
B 1320
C 1615
D 1855
Correct Answer
Option A
Solution

Given,

∑k=131(31Ck)(31Ck−1)−∑k=130(30Ck)(30Ck−1)=α(60!)(30!)(31!)\sum\limits_{k = 1}^{31} {\left( {{}^{31}{C_k}} \right)\left( {{}^{31}{C_{k - 1}}} \right) - \sum\limits_{k = 1}^{30} {\left( {{}^{30}{C_k}} \right)\left( {{}^{30}{C_{k - 1}}} \right) = {{\alpha (60!)} \over {(30!)(31!)}}} }

Now,

∑k=131(31Ck)(31Ck−1)\sum\limits_{k = 1}^{31} {\left( {{}^{31}{C_k}} \right)\left( {{}^{31}{C_{k - 1}}} \right)}
=(31C1 . 31C0+31C2 . 31C1+31C3 . 31C2+  ......  +  31C31 . 31C30)= \left( {{}^{31}{C_1}\,.\,{}^{31}{C_0} + {}^{31}{C_2}\,.\,{}^{31}{C_1} + {}^{31}{C_3}\,.\,{}^{31}{C_2} + \,\,......\,\, + \,\,{}^{31}{C_{31}}\,.\,{}^{31}{C_{30}}} \right)
=(31C0 . 31C31−1+31C1 . 31C31−2+  .....  +  31C30 . 31C31−31)= \left( {{}^{31}{C_0}\,.\,{}^{31}{C_{31 - 1}} + {}^{31}{C_1}\,.\,{}^{31}{C_{31 - 2}} + \,\,.....\,\, + \,\,{}^{31}{C_{30}}\,.\,{}^{31}{C_{31 - 31}}} \right)

[using

nCr=nCn−r{}^n{C_r} = {}^n{C_{n - r}}

]

=(31C0 . 31C30+31C1 . 31C29+  .....  +  31C30 . 31C0)= \left( {{}^{31}{C_0}\,.\,{}^{31}{C_{30}} + {}^{31}{C_1}\,.\,{}^{31}{C_{29}} + \,\,.....\,\, + \,\,{}^{31}{C_{30}}\,.\,{}^{31}{C_0}} \right)
=62C30= {}^{62}{C_{30}}

Now,

∑k=13030Ck . 30Ck−1\sum\limits_{k = 1}^{30} {{}^{30}{C_k}\,.\,{}^{30}{C_{k - 1}}}
=(30C1 . 30C0+30C2 . 30C1+  .....  +  30C30 . 30C29)= \left( {{}^{30}{C_1}\,.\,{}^{30}{C_0} + {}^{30}{C_2}\,.\,{}^{30}{C_1} + \,\,.....\,\, + \,\,{}^{30}{C_{30}}\,.\,{}^{30}{C_{29}}} \right)
=(30C0 . 30C29+30C1 . 30C28+  ....  +  30C29 . 30C0)= \left( {{}^{30}{C_0}\,.\,{}^{30}{C_{29}} + {}^{30}{C_1}\,.\,{}^{30}{C_{28}} + \,\,....\,\, + \,\,{}^{30}{C_{29}}\,.\,{}^{30}{C_0}} \right)
=60C29= {}^{60}{C_{29}}

∴\therefore

60C30−60C29=α(60!)30! 31!{}^{60}{C_{30}} - {}^{60}{C_{29}} = {{\alpha (60!)} \over {30!\,31!}}
⇒62 . 61 . 60!30! 32!−60!29! 31!=α(60!)30! 31!\Rightarrow {{62\,.\,61\,.\,60!} \over {30!\,32!}} - {{60!} \over {29!\,31!}} = {{\alpha (60!)} \over {30!\,31!}}
⇒62 . 61 . 60!30! 32!−60!30!30 . 31!=α(60!)30! 31!\Rightarrow {{62\,.\,61\,.\,60!} \over {30!\,32!}} - {{60!} \over {{{30!} \over {30}}\,.\,31!}} = {{\alpha (60!)} \over {30!\,31!}}
⇒60!30! 31!(62 . 6132−30)=α(60!)30! 31!\Rightarrow {{60!} \over {30!\,31!}}\left( {{{62\,.\,61} \over {32}} - 30} \right) = {{\alpha (60!)} \over {30!\,31!}}
⇒α=62 . 6132−30\Rightarrow \alpha = {{62\,.\,61} \over {32}} - 30
⇒16α=62 . 61−30×322\Rightarrow 16\alpha = {{62\,.\,61 - 30 \times 32} \over 2}
⇒16α=28222=1411\Rightarrow 16\alpha = {{2822} \over 2} = 1411
Q48
The remainder when (2021)2023 is divided by 7 is :
A 1
B 2
C 5
D 6
Correct Answer
Option C
Solution

(2021)2023 = (2016 + 5)2023 [here 2016 is divisible by 7] = 2023C0 (2016)2023 + ..........

+ 2023C2022 (2016) (5)2022 + 2023C2023 (5)2023 = 2016 [2023C0 .

(2016)2022 + .......

+ 2023C2022 .

(5)2022] + (5)2023 = 2016λ\lambda + (5)2023 = 7 ×\times 288λ\lambda + (5)2023 = 7K + (5)2023 ......

(1) Now, (5)2023 = (5)2022 .

5 = (53)674 .

5 = (125)674 .

5 = (126 −- 1)674 .

5 = 5[674C0 (126)674 + .........

−- 674C673 (126) + 674C674] = 5 ×\times 126 [674C0(126)673 + .......

−- 674C673] + 5 = 5 .

7 .

18 [674C0(126)673 + .......

−- 674C673] + 5 = 7λ\lambda + 5 Replacing (5)2023 in equation (1) with 7λ\lambda + 5, we get, (2021)2023 = 7K + 7λ\lambda + 5 = 7(K + λ\lambda) + 5 ∴\therefore Remainer = 5

Q49
The coefficient of x101 in the expression (5+x)500+x(5+x)499+x2(5+x)498+  .....  +  x500{(5 + x)^{500}} + x{(5 + x)^{499}} + {x^2}{(5 + x)^{498}} + \,\,.....\,\, + \,\,{x^{500}}, x > 0, is
A 501C101 (5)399
B 501C101 (5)400
C 501C100 (5)400
D 500C101 (5)399
Correct Answer
Option A
Solution

Given,

(5+x)500+x(5+x)499+x2(5+x)498  +{(5 + x)^{500}} + x{(5 + x)^{499}} + {x^2}{(5 + x)^{498}}\,\, +

......

x500{x^{500}}

This is a G.P. with first term

(5+x)500{(5 + x)^{500}}

Common ratio

=x(5+x)499(5+x)500=x5+x= {{x{{(5 + x)}^{499}}} \over {{{(5 + x)}^{500}}}} = {x \over {5 + x}}

and 501 terms present. ∴\therefore Sum

=(5+x)500((x5+x)501−1)x5+x−1= {{{{(5 + x)}^{500}}\left( {{{\left( {{x \over {5 + x}}} \right)}^{501}} - 1} \right)} \over {{x \over {5 + x}} - 1}}
=(5+x)500(x501−(5+x)501(5+x)501)x−5−x5+x= {{{{(5 + x)}^{500}}\left( {{{{x^{501}} - {{(5 + x)}^{501}}} \over {{{(5 + x)}^{501}}}}} \right)} \over {{{x - 5 - x} \over {5 + x}}}}
=x501−(5+x)5015+x−55+x= {{{{{x^{501}} - {{(5 + x)}^{501}}} \over {5 + x}}} \over {{{ - 5} \over {5 + x}}}}
=15((5+x)501−x501)= {1 \over 5}\left( {{{\left( {5 + x} \right)}^{501}} - {x^{501}}} \right)

Coefficient of x101 in

(5+x)501{(5 + x)^{501}}

is

=501C101 . 5400= {}^{501}{C_{101}}\,.\,{5^{400}}

∴\therefore In

15((5+x)500−x501){1 \over 5}\left( {{{(5 + x)}^{500}} - {x^{501}}} \right)

coefficient of x101 is

=15 . 501C101 . 5400= {1 \over 5}\,.\,{}^{501}{C_{101}}\,.\,{5^{400}}
=501C101 . 5399= {}^{501}{C_{101}}\,.\,{5^{399}}
Q50
If 12⋅(15C1)+22⋅(15C2)+32⋅(15C3)+…+152⋅(15C15)=2m⋅3n⋅5k1^2 \cdot\left({ }^{15} C_1\right)+2^2 \cdot\left({ }^{15} C_2\right)+3^2 \cdot\left({ }^{15} C_3\right)+\ldots+15^2 \cdot\left({ }^{15} C_{15}\right)=2^m \cdot 3^n \cdot 5^k, where m,n,k∈Nm, n, k \in \mathbf{N}, then m+n+k\mathrm{m}+\mathrm{n}+\mathrm{k} is equal to :
A 20
B 19
C 18
D 21
Correct Answer
Option B
Solution

∑r=115r2⋅15Cr(rnCr=nn−1Cr−1)\sum_{r=1}^{15} r^2 \cdot{ }^{15} C_r \quad\left(r^n C_r=n^{n-1} C_{r-1}\right)

=15∑r=115r⋅14Cr−1=15∑r=115(r−1+1)14Cr−1=15⋅∑r=115(r−1)14Cr−1+15⋅∑r=11514Cr−1=15⋅14⋅213+15⋅214=15⋅214(7+1)=5⋅3⋅217n+m+k=17+1+1=19\begin{aligned} & =15 \sum_{r=1}^{15} r \cdot{ }^{14} C_{r-1} \\ & =15 \sum_{r=1}^{15}(r-1+1){ }^{14} C_{r-1} \\ & =15 \cdot \sum_{r=1}^{15}(r-1){ }^{14} C_{r-1}+15 \cdot \sum_{r=1}^{15}{ }^{14} C_{r-1} \\ & =15 \cdot 14 \cdot 2^{13}+15 \cdot 2^{14} \\ & =15 \cdot 2^{14}(7+1) \\ & =5 \cdot 3 \cdot 2^{17} \\ & n+m+k=17+1+1=19 \end{aligned}
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