Binomial Theorem Questions — JEE Mathematics

Q1

In the binomial expansion of

{\left( {a - b} \right)^n},\,\,\,n \ge 5,

the sum of

{5^{th}}

and

{6^{th}}

terms is zero, then

a/b

equals

A

{{n - 5} \over 6}

B

{{n - 4} \over 5}

C

{5 \over {n - 4}}

D

{6 \over {n - 5}}

Correct Answer

Option B

Solution

According to the question, t5 + t6 = 0 $\therefore$

{}^n{C_4}.{a^{n - 4}}.{b^4}

+

\left( { - {}^n{C_5}.{a^{n - 5}}.{b^5}} \right)

= 0 By solving we get,

{a \over b} = {{n - 4} \over 5}

Q2

The coefficient of

{x^7}

in the expansion of

{\left( {1 - x - {x^2} + {x^3}} \right)^6}

is

A

-132

B

-144

C

132

D

144

Correct Answer

Option B

Solution

Given,

{\left( {1 - x - {x^2} + {x^3}} \right)^6}

=

{\left[ {\left( {1 - x} \right) - {x^2}\left( {1 - x} \right)} \right]^6}

=

{\left( {1 - x} \right)^6}{\left( {1 - {x^2}} \right)^6}

=

\left( {1 + {}^6{C_1}( - x) + {}^6{C_2}{{( - x)}^2} + {}^6{C_3}{{( - x)}^3} + .......} \right)\times

\left( {1 + {}^6{C_1}( - {x^2}) + {}^6{C_2}{{( - {x^2})}^2} + {}^6{C_3}{{( - {x^2})}^3} + .......} \right)

$\therefore$ Coefficient of x7 =

- {}^6{C_1} \times - {}^6{C_3} + \left( { - {}^6{C_3}} \right) \times {}^6{C_2} + \left( { - {}^6{C_5}} \right) \times - {}^6{C_1}

= 120 - 300 + 36 = - 144

Q3

If the number of terms in the expansion of

{\left( {1 - {2 \over x} + {4 \over {{x^2}}}} \right)^n},\,x \ne 0,

is 28, then the sum of the coefficients of all the terms in this expansion, is :

A 243

B 729

C 64

D 2187

Correct Answer

Option B

Solution

Total no of terms in

{\left( {1 - {2 \over x} + {4 \over {{x^2}}}} \right)^n}

=

{}^{n + 2}{C_2}

= 28 (n+2)(n+1) = 56

\Rightarrow n = 6

Sum of coefficient = (1 - 2 + 4)6 = 36 = 729

Q4

If the fourth term in the binomial expansion of

{\left( {{2 \over x} + {x^{{{\log }_8}x}}} \right)^6}

(x > 0) is 20 × 87, then a value of x is :

A 8–2

B 82

C 83

D 8

Correct Answer

Option B

Solution

{\left( {{2 \over x} + {x^{{{\log }_8}x}}} \right)^6}

Given T4 = 20 × 87 $\Rightarrow$

{}^6{C_3}{\left( {{2 \over x}} \right)^3}{\left( {{x^{{{\log }_8}x}}} \right)^3}

= 20 × 87 $\Rightarrow$ 20 $\times$

{8 \over {{x^3}}} \times {x^{3{{\log }_8}x}}

= 20 × 87 $\Rightarrow$

{x^{3{{\log }_8}x - 3}}

= 86 Taking

{{{\log }_8}}

both side $\Rightarrow$ (

{3{{\log }_8}x - 3}

) $\times$

{{{\log }_8}x}

= 6 $\Rightarrow$

3{\left( {{{\log }_8}x} \right)^2}

- 3

{{{\log }_8}x}

= 6 $\Rightarrow$

{\left( {{{\log }_8}x} \right)^2}

-

{{{\log }_8}x}

= 2 $\Rightarrow$ (

{{{\log }_8}x}

- 2)(

{{{\log }_8}x}

+ 1) = 0 $\Rightarrow$

{{{\log }_8}x}

= 2 or

{{{\log }_8}x}

= -1 $\Rightarrow$ x = 82 or x =

{1 \over 8}

Q5

The coefficient of

x^{5}

in the expansion of

\left(2 x^{3}-\dfrac{1}{3 x^{2}}\right)^{5}

is :

A

\dfrac{26}{3}

B

\dfrac{80}{9}

C 9

D 8

Correct Answer

Option B

Solution

Given,

\left(2 x^{3}-\frac{1}{3 x^{2}}\right)^{5}

General term,

\begin{aligned} & T_{r+1}={ }^5 C_r\left(2 x^3\right)^{5-r}\left(\frac{-1}{3 x^2}\right)^r={ }^5 C_r \frac{(2)^{5-r}}{(-3)^r}(x)^{15-5 r} \\\\ & \therefore 15-5 \mathrm{r}=5 \\\\ & \therefore \mathrm{r}=2 \\\\ & T_3=10\left(\frac{8}{9}\right) x^5 \end{aligned}

So, coefficient is $\dfrac{80}{9}$ .

Q6

The value of

\sum\limits_{r = 0}^{20} {{}^{50 - r}{C_6}}

is equal to:

A

{}^{50}{C_6} - {}^{30}{C_6}

B

{}^{51}{C_7} - {}^{30}{C_7}

C

{}^{50}{C_7} - {}^{30}{C_7}

D

{}^{51}{C_7} + {}^{30}{C_7}

Correct Answer

Option B

Solution

\sum\limits_{r = 0}^{20} {} {}^{50 - r}{C_6} = {}^{50}{C_6} + {}^{49}{C_6} + {}^{48}{C_6} + .... + {}^{30}{C_6}

= {}^{50}{C_6} + {}^{49}{C_6} + .... + {}^{31}{C_6} + ({}^{30}{C_6} + {}^{30}{C_7}) - {}^{30}{C_7}

= {}^{50}{C_6} + {}^{49}{C_6} + .... + ({}^{31}{C_6} + {}^{31}{C_7}) - {}^{30}{C_7}

= {}^{50}{C_6} + {}^{50}{C_7} - {}^{30}{C_7}

= {}^{51}{C_7} - {}^{30}{C_7}

[As

{{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}}

]

Q7

r

and

n

are positive integers

\,r > 1,\,n > 2

and coefficient of

\,{\left( {r + 2} \right)^{th}}

term and

3{r^{th}}

term in the expansion of

{\left( {1 + x} \right)^{2n}}

are equal, then

n

equals

A

3r

B

3r + 1

C

2r

D

2r + 1

Correct Answer

Option C

Solution

\,{\left( {r + 2} \right)^{th}}

term =

{}^{2n}{C_{r+1}}{\left( x \right)^r}

And coefficient of

\,{\left( {r + 2} \right)^{th}}

=

{}^{2n}{C_{r+1}}

3{r^{th}}

term =

{}^{2n}{C_{3r - 1}}{\left( x \right)^{3r - 1}}

And coefficient of

3{r^{th}}

term =

{}^{2n}{C_{3r - 1}}

According to the question,

{}^{2n}{C_{r+1}}

=

{}^{2n}{C_{3r - 1}}

\Rightarrow \left( {r + 1} \right) + \left( {3r - 1} \right) = 2n

[As if

{}^n{C_p} = {}^n{C_q}

then p + q = n]

\Rightarrow 4r = 2n

\Rightarrow n = 2r

Q8

The coefficients of

{x^p}

and

{x^q}

in the expansion of

{\left( {1 + x} \right)^{p + q}}

are

A equal

B equal with opposite signs

C reciprocals of each other

D none of these

Correct Answer

Option A

Solution

Here in this expansion

{\left( {1 + x} \right)^{p + q}}

The general term =

{T_{r + 1}} = {}^{p + q}{C_r}.{\left( x \right)^r}

$\therefore$

{x^p}

will be present in the term =

{}^{p + q}{C_p}.{\left( x \right)^p}

So coefficient of

{x^p}

=

{}^{p + q}{C_p}

And

{x^q}

will be present in the term =

{}^{p + q}{C_q}.{\left( x \right)^q}

$\therefore$ coefficient of

{x^q}

=

{}^{p + q}{C_q}

We know

{}^n{C_r}

=

{}^n{C_{n - r}}

$\therefore$

{}^{p + q}{C_q}

=

{}^{p + q}{C_{\left( {p + q} \right) - q}}

=

{}^{p + q}{C_p}

So coefficients of

{x^p}

and

{x^q}

are equal.

Q9

If

x

is positive, the first negative term in the expansion of

{\left( {1 + x} \right)^{27/5}}

is

A 6th term

B 7th term

C 5th term

D 8th term.

Correct Answer

Option D

Solution

General term of

{\left( {1 + x} \right)^{n}}

is (

{T_{r + 1}}

) =

{{n\left( {n - 1} \right).....\left( {n - r + 1} \right)} \over {1.2.3....r}}{x^r}

$\therefore$ General term of

{\left( {1 + x} \right)^{27/5}}

=

{{{{27} \over 5}\left( {{{27} \over 5} - 1} \right).....\left( {{{27} \over 5} - r + 1} \right)} \over {1.2.3....r}}{x^r}

For first negative term,

{\left( {{{27} \over 5} - r + 1} \right)}

< 0

\Rightarrow r > {{27} \over 5} + 1

\Rightarrow r > {{32} \over 5}

\Rightarrow r > 6.4

$\therefore$ r = 7

{T_{7 + 1}} = {T_8}

means 8th term is the first negative term.

Q10

25^{190}-19^{190}-8^{190}+2^{190}

is divisible by :

A 14 but not by 34

B neither 14 nor 34

C both 14 and 34

D 34 but not by 14

Correct Answer

Option D

Solution

The given expression is divisible by 6 and 17 .

Also, $25^{190}-8^{190}$ is not divisible by 7 but $19^{190}-2^{190}$ is divisible by 7 , So, $25^{190}-19^{190}-8^{190}+2^{190}$ is divisible by 34 but not by 14 .