Binomial Theorem — Page 4 | JEE Mathematics

Q31

The value of -15C1 + 2.15C2 – 3.15C3 + ... - 15.15C15 + 14C1 + 14C3 + 14C5 + ...+ 14C11 is :

A 213 - 13

B 216 - 1

C 214

D 213 - 14

Correct Answer

Option D

Solution

- {}^{15}{C_1} + 2.{}^{15}{C_2} - 3.{}^{15}{C_3} + ....\,. - 15.{}^{15}{C_{15}}

= \sum\limits_{r = 1}^{15} {{{( - 1)}^r}.r.{}^{15}{C_r}}

= \sum\limits_{r = 1}^{15} {{{( - 1)}^2}.r.{{15} \over r}} .{}^{14}{C_{r - 1}}

= 15\sum\limits_{r = 1}^{15} {{{( - 1)}^2}.{}^{14}{C_{r - 1}}}

= 15\left( { - {}^{14}{C_0} + {}^{14}{C_1} - {}^{14}{C_2} + .... - {}^{14}{C_{14}}} \right)

= 15(0) = 0

We know, $\Rightarrow$ 214 - 1

= {}^{14}{C_1} + {}^{14}{C_3} + {}^{14}{C_5} .... + {}^{14}{C_{13}}

$\Rightarrow$ 213

= {}^{14}{C_1} + {}^{14}{C_3} + {}^{14}{C_5} .... + {}^{14}{C_{13}}

Also let, S = 14C1 + 14C3 + 14C5 + ...

+ 14C11 $\Rightarrow$ S + 14C13 = 14C1 + 14C3 + 14C5 + ...

+ 14C11 + 14C13 $\Rightarrow$ S + 14C13 = 213 $\Rightarrow$ S + 14 = 213 $\Rightarrow$ S = 213 - 14 Other Method : We know,

{(1 - x)^{15}} = {}^{15}{C_0} - {}^{15}{C_1}x + {}^{15}{C_2}{x^2} - ..... - {}^{15}{C_{15}}{x^{15}}

Differentiating both sides with respect to x,

15{(1 - x)^{14}}( - 1) = - {}^{15}{C_1} + 2{}^{15}{C_2}x - 3{}^{15}{C_3}{x^2} + ....... - 15{}^{15}{C_{15}}{x^{14}}

Put

x = 1

\Rightarrow 0 = - {}^{15}{C_1} + 2{}^{15}{C_2} - 3{}^{15}{C_3} + .... - 15{}^{15}{C_{15}}

We know, $\Rightarrow$ 214 - 1

= {}^{14}{C_1} + {}^{14}{C_3} + {}^{14}{C_5} .... + {}^{14}{C_{13}}

$\Rightarrow$ 213

= {}^{14}{C_1} + {}^{14}{C_3} + {}^{14}{C_5} .... + {}^{14}{C_{13}}

Also let, S = 14C1 + 14C3 + 14C5 + ...

+ 14C11 $\Rightarrow$ S + 14C13 = 14C1 + 14C3 + 14C5 + ...

+ 14C11 + 14C13 $\Rightarrow$ S + 14C13 = 213 $\Rightarrow$ S + 14 = 213 $\Rightarrow$ S = 213 - 14

Q32

The coefficient of x7 in the expression (1 + x)10 + x(1 + x)9 + x2(1 + x)8 + ......+ x10 is:

A 120

B 330

C 420

D 210

Correct Answer

Option B

Solution

(1 + x)10 + x(1 + x)9 + x2(1 + x)8 + ......+ x10 This is a G.P where First term, a = (1 + x)10 common ratio, r =

{x \over {1 + x}}

Number of terms = 11 Sum of G.P =

{{{{\left( {1 + x} \right)}^{10}}\left( {1 - {{\left( {{x \over {1 + x}}} \right)}^{11}}} \right)} \over {1 - {x \over {1 + x}}}}

= (1 + x)11 – x11 So Coefficient of x7 is 11C7 = 330

Q33

If the number of integral terms in the expansion of (31/2 + 51/8)n is exactly 33, then the least value of n is :

A 264

B 256

C 128

D 248

Correct Answer

Option B

Solution

General term of the expression,

{T_{r + 1}} = {}^n{C_r}{\left( {{3^{{1 \over 2}}}} \right)^{n - r}}{\left( {{5^{{1 \over 8}}}} \right)^r}

= {}^n{C_r}{\left( 3 \right)^{{{n - r} \over 2}}}{\left( 5 \right)^{{r \over 8}}}

We will get integral term when

{{n - r} \over 2}

and

{r \over 8}

are integer $\therefore$ (1) n $-$ r is multiple of 2 $\Rightarrow$ n $-$ r = 0, 2, 4, ...... (2) r is multiple of 8 $\Rightarrow$ r = 0, 8, 16, .......

From this two conditions common values are = 0, 8, 16, ....... which will becomes integral terms.

Given that there are 33 integral terms.

Here first integral term at 0th position.

Second integral term at 8th position. $\therefore$ 33th integral term will be at = 0 + (33 $-$ 1)8 = 256 So, there should be at least 256 terms.

Q34

If

n \ge 2

is a positive integer, then the sum of the series

{}^{n + 1}{C_2} + 2\left( {{}^2{C_2} + {}^3{C_2} + {}^4{C_2} + ... + {}^n{C_2}} \right)

is :

A

{{n(2n + 1)(3n + 1)} \over 6}

B

{{n(n + 1)(2n + 1)} \over 6}

C

{{n{{(n + 1)}^2}(n + 2)} \over {12}}

D

{{n(n - 1)(2n + 1)} \over 6}

Correct Answer

Option B

Solution

{}^{n + 1}{C_2} + 2\left( {{}^2{C_2} + {}^3{C_2} + {}^4{C_2} + ........ + {}^n{C_2}} \right)

{}^{n + 1}{C_2} + 2\left( {{}^3{C_2} + {}^3{C_2} + {}^4{C_2} + ........ + {}^n{C_2}} \right)

use

\left\{ {{}^n{C_{r + 1}} + {}^n{C_r} = {}^{n + 1}{C_r}} \right\}

= {}^{n + 1}{C_2} + 2\left( {{}^4{C_3} + {}^4{C_2} + {}^5{C_3} + ........ + {}^n{C_2}} \right)

= {}^{n + 1}{C_2} + 2\left( {{}^5{C_3} + {}^5{C_2} + ........ + {}^n{C_2}} \right)

\begin{aligned}& . \\ & . \\ & . \\ & . \\ & .\end{aligned}

= {}^{n + 1}{C_2} + 2\left( {{}^n{C_3} + {}^n{C_2}} \right)

= {}^{n + 1}{C_2} + 2.{}^{n + 1}{C_3}

= {{(n + 1)n} \over 2} + 2.{{(n + 1)(n)(n - 1)} \over {2.3}}

= {{n(n + 1)(2n + 1)} \over 6}

Q35

The maximum value of the term independent of 't' in the expansion of

{\left( {t{x^{{1 \over 5}}} + {{{{(1 - x)}^{{1 \over {10}}}}} \over t}} \right)^{10}}

where x

\in

(0, 1) is :

A

{{10!} \over {\sqrt 3 {{(5!)}^2}}}

B

{{2.10!} \over {3\sqrt 3 {{(5!)}^2}}}

C

{{10!} \over {3{{(5!)}^2}}}

D

{{2.10!} \over {3{{(5!)}^2}}}

Correct Answer

Option B

Solution

{T_{r + 1}} = {}^{10}{C_r}{(t{x^{1/5}})^{10 - r}}{\left[ {{{{{(1 - x)}^{1/10}}} \over t}} \right]^r}

= {}^{10}{C_r}{t^{(10 - 2r)}} \times {x^{{{10 - r} \over 5}}} \times {(1 - x)^{{r \over {10}}}}

\Rightarrow 10 - 2r = 0 \Rightarrow r = 5

$\therefore$

{T_6} = {}^{10}{C_5} \times x\sqrt {1 - x}

At maximum,

{{d{T_6}} \over {dx}} = {}^{10}{C_5}\left[ {\sqrt {1 - x} - {x \over {2\sqrt {1 - x} }}} \right] = 0

$\Rightarrow$

1 - x = x/2 \Rightarrow 3x = 2 \Rightarrow x = 2/3

{T_6}{|_{\max }} = {{10!} \over {5!5!}} \times {2 \over {3\sqrt 3 }}

=

{{2.10!} \over {3\sqrt 3 {{(5!)}^2}}}

Q36

If

{{}^{20}{C_r}}

is the co-efficient of xr in the expansion of (1 + x)20, then the value of

\sum\limits_{r = 0}^{20} {{r^2}.{}^{20}{C_r}}

is equal to :

A

420 \times {2^{19}}

B

380 \times {2^{19}}

C

380 \times {2^{18}}

D

420 \times {2^{18}}

Correct Answer

Option D

Solution

\sum\limits_{r = 0}^{20} {{r^2}.{}^{20}{C_r}}

\sum {(4(r - 1) + r).{}^{20}{C_r}}

\sum {r(r - 1).{{20 \times 19} \over {r(r - 1)}}} .{}^{18}{C_r} + r.{{20} \over r}.\sum {{}^{19}{C_{r - 1}}}

\Rightarrow 20 \times {19.2^{18}} + {20.2^{19}}

\Rightarrow 420 \times {2^{18}}

Q37

Let [ x ] denote greatest integer less than or equal to x. If for n

\in

N,

{(1 - x + {x^3})^n} = \sum\limits_{j = 0}^{3n} {{a_j}{x^j}}

, then

\sum\limits_{j = 0}^{\left[ {{{3n} \over 2}} \right]} {{a_{2j}} + 4} \sum\limits_{j = 0}^{\left[ {{{3n - 1} \over 2}} \right]} {{a_{2j}} + 1}

is equal to :

A 2n

-

1

B n

C 2

D 1

Correct Answer

Option C

Solution

{(1 - x + {x^3})^n} = \sum\limits_{j = 0}^{3n} {{a_j}{x^j}}

(1 - x + {x^3}) = {a_0} + {a_1}x + {a_2}{x^2} + ...... + {a_{3n}}{x^{3n}}

Put x = 1

1 = {a_0} + {a_1} + {a_2} + {a_3} + {a_4} + ........ + {a_{3n}}

...... (1) Put x = $-$ 1

1 = {a_0} - {a_1} + {a_2} - {a_3} + {a_4} + ........( - 1){}^{3n}{a_{3n}}

..... (2) Add (1) + (2)

\Rightarrow {a_0} + {a_2} + {a_4} + {a_6} + ...... = 1

Sub (1) $-$ (2)

\Rightarrow {a_1} + {a_3} + {a_5} + {a_7} + ...... = 0

Now,

\sum\limits_{j = 0}^{\left[ {{{3n} \over 2}} \right]} {{a_{2j}}} + 4\sum\limits_{j = 0}^{\left[ {{{3n - 1} \over 2}} \right]} {{a_{2j }}} + 1

= ({a_0} + {a_2} + {a_4} + ......) + 4({a_1} + {a_3} + .....)

= 1 + 4 \times 0

+ 1

= 1 + 1 = 2

Q38

Let (1 + x + 2x2)20 = a0 + a1x + a2x2 + .... + a40x40. Then a1 + a3 + a5 + ..... + a37 is equal to

A 220(220

-

21)

B 219(220

-

21)

C 219(220

+

21)

D 220(220

+

21)

Correct Answer

Option B

Solution

{(1 + x + 2{x^2})^{20}} = {a_0} + {a_1}x + {a_2}{x^2} + .... + {a_{40}}{x^{40}}

Put x = 1

\Rightarrow {4^{20}} = {a_0} + {a_1} + ....... + {a_{40}}

..... (i) Put x = $-$ 1

\Rightarrow {2^{20}} = {a_0} - {a_1} + ....... + - {a_{39}} + {a_{40}}

..... (ii) by (i) $-$ (ii) we get,

{4^{20}} - {2^{20}} = 2({a_1} + {a_3} + ...... + {a_{37}} + {a_{39}})

\Rightarrow {a_1} + {a_3} + ...... + {a_{37}} = {2^{39}} - {2^{19}} - {a_{39}}

..... (iii) a39 = coeff. x39 in (1 + x + 2x2)20

= {{20!} \over {0!1!9!}}{(1)^0}{(1)^1}{(2)^{19}}

= {20.2^{19}}

$\therefore$

{a_1} + {a_3} + ...... + {a_{37}} = {2^{39}} - {2^{19}}.21

\Rightarrow {2^{19}}({2^{20}} - 21)

Q39

The coefficient of x256 in the expansion of (1

-

x)101 (x2 + x + 1)100 is :

A

{}^{100}{C_{16}}

B

{}^{100}{C_{15}}

C

-

{}^{100}{C_{16}}

D

-

{}^{100}{C_{15}}

Correct Answer

Option B

Solution

{(1 - x)^{101}}{({x^2} + x + 1)^{100}}

Coefficient of

{x^{256}} = {[(1 - x)(1 + x + {x^2})]^{100}}(1 - x) = {(1 - {x^3})^{100}}(1 - x)

\Rightarrow ({}^{100}{C_0} - {}^{100}{C_1}{x^3} + {}^{100}{C_2}{x^6} - {}^{100}{C_3}{x^9}...)(1 - x)

\sum {{{( - 1)}^r}{}^{100}{C_r}{x^{3r}}(1 - x)}

\Rightarrow 3r = 256

or

255 \Rightarrow r = {{256} \over 3}

(Reject) r = 85 Coefficient =

{}^{100}{C_{85}} = {}^{100}{C_{15}}

Q40

If b is very small as compared to the value of a, so that the cube and other higher powers of

{b \over a}

can be neglected in the identity

{1 \over {a - b}} + {1 \over {a - 2b}} + {1 \over {a - 3b}} + ..... + {1 \over {a - nb}} = \alpha n + \beta {n^2} + \gamma {n^3}

, then the value of

\gamma

is :

A

{{{a^2} + b} \over {3{a^3}}}

B

{{a + b} \over {3{a^2}}}

C

{{{b^2}} \over {3{a^3}}}

D

{{a + {b^2}} \over {3{a^3}}}

Correct Answer

Option C

Solution

{(a - b)^{ - 1}} + {(a - 2b)^{ - 1}} + .... + {(a - nb)^{ - 1}}

= {1 \over a}\sum\limits_{r = 1}^n {{{\left( {1 - {{rb} \over a}} \right)}^{ - 1}}}

= {1 \over a}\sum\limits_{r = 1}^n {\left\{ {\left( {1 + {{rb} \over a} + {{{r^2}{b^2}} \over {{a^2}}}} \right) + (terms\,to\,be\,neglected)} \right\}}

= {1 \over a}\left[ {n + {{n(n + 1)} \over 2}.{b \over a} + {{n(n + 1)(2n + 1)} \over 6}.{{{b^2}} \over {{a^2}}}} \right]

= {1 \over a}\left[ {{n^3}\left( {{{{b^2}} \over {3{a^2}}}} \right) + .....} \right]

So,

\gamma = {{{b^2}} \over {3{a^3}}}