Let K be the sum of the coefficients of the odd powers of x in the expansion of (1+x)^99 . Let a be the middle term in the expansion of ( 2 + 1 \over \sqrt 2 )^200 . If ^200C_99K \over a = 2^lm \over n , where m and n are odd numbers, then the ordered pair (l,n) is equal to

Binomial Theorem PYQ — JEE Main 2023 · 29 Jan · Shift 2 · Q35 with Solution

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Correct answer: A

Find $K$

The sum of coefficients of odd powers of $x$ in $(1+x)^{99}$ is given by

\frac{(1+1)^{99}-(1-1)^{99}}{2} = \frac{2^{99}-0}{2}=2^{98}.

So, $K=2^{98}.$

Find the middle term $a$ in $\left(2+\frac{1}{\sqrt2}\right)^{200}$

For $(u+v)^{200}$ , the number of terms is $201$ , so the middle term is the $\left(\frac{200}{2}+1\right)=101$ -th term.

General term:

T_{r+1}=\binom{200}{r}2^{200-r}\left(\frac{1}{\sqrt2}\right)^r.

For the middle term, $r=100$ :

a=T_{101}=\binom{200}{100}2^{100}\left(\frac{1}{\sqrt2}\right)^{100}.

Now,

\left(\frac{1}{\sqrt2}\right)^{100}=2^{-50}.

Hence,

a=\binom{200}{100}2^{100}2^{-50}=\binom{200}{100}2^{50}.

Compute $\displaystyle \frac{{}^{200}C_{99}\,K}{a}$

Substitute $K=2^{98}$ and $a=\binom{200}{100}2^{50}$ :

\frac{\binom{200}{99}2^{98}}{\binom{200}{100}2^{50}} =2^{48}\cdot \frac{\binom{200}{99}}{\binom{200}{100}}.

Now use

\frac{\binom{200}{99}}{\binom{200}{100}}= rac{100!100!}{99!101!}=\frac{100}{101}.

Therefore,

\frac{{}^{200}C_{99}K}{a}=2^{48}\cdot \frac{100}{101} =2^{48}\cdot \frac{2^2\cdot 25}{101} =\frac{2^{50}\cdot 25}{101}.

So in the form

\frac{2^l m}{n},

where $m,n$ are odd, we get

l=50,\qquad n=101.

Thus the ordered pair is

(l,n)=(50,101).

Check with options

This matches Option A.

Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So they agree.

More from Binomial Theorem

View written solutionFree

Correct answer: A

Find $K$

The sum of coefficients of odd powers of $x$ in $(1+x)^{99}$ is given by

\frac{(1+1)^{99}-(1-1)^{99}}{2} = \frac{2^{99}-0}{2}=2^{98}.

So, $K=2^{98}.$

Find the middle term $a$ in $\left(2+\frac{1}{\sqrt2}\right)^{200}$

For $(u+v)^{200}$ , the number of terms is $201$ , so the middle term is the $\left(\frac{200}{2}+1\right)=101$ -th term.

General term:

T_{r+1}=\binom{200}{r}2^{200-r}\left(\frac{1}{\sqrt2}\right)^r.

For the middle term, $r=100$ :

a=T_{101}=\binom{200}{100}2^{100}\left(\frac{1}{\sqrt2}\right)^{100}.

Now,

\left(\frac{1}{\sqrt2}\right)^{100}=2^{-50}.

Hence,

a=\binom{200}{100}2^{100}2^{-50}=\binom{200}{100}2^{50}.

Compute $\displaystyle \frac{{}^{200}C_{99}\,K}{a}$

Substitute $K=2^{98}$ and $a=\binom{200}{100}2^{50}$ :

\frac{\binom{200}{99}2^{98}}{\binom{200}{100}2^{50}} =2^{48}\cdot \frac{\binom{200}{99}}{\binom{200}{100}}.

Now use

\frac{\binom{200}{99}}{\binom{200}{100}}= rac{100!100!}{99!101!}=\frac{100}{101}.

Therefore,

\frac{{}^{200}C_{99}K}{a}=2^{48}\cdot \frac{100}{101} =2^{48}\cdot \frac{2^2\cdot 25}{101} =\frac{2^{50}\cdot 25}{101}.

So in the form

\frac{2^l m}{n},

where $m,n$ are odd, we get

l=50,\qquad n=101.

Thus the ordered pair is

(l,n)=(50,101).

Check with options

This matches Option A.

Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So they agree.

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