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MCQMediumJEE 2023Middle Term

JEE Mathematics 2023 Question with Solution

Let KK be the sum of the coefficients of the odd powers of xx in the expansion of (1+x)99(1+x)^{99}. Let aa be the middle term in the expansion of (2+12)200\left(2 + \frac{1}{\sqrt{2}}\right)^{200}. If 200C99Ka=2ℓmn,\frac{^{200}C_{99}K}{a} = \frac{2^\ell m}{n}, where mm and nn are odd numbers, then the ordered pair (ℓ,n)(\ell, n) is equal to:

  • A

    (50,51)(50, 51)

  • B

    (51,99)(51, 99)

  • C

    (50,101)(50, 101)

  • D

    (51,101)(51, 101)

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: KK is the sum of coefficients of odd powers of xx in (1+x)99(1+x)^{99}. Also, aa is the middle term of (2+12)200\left(2 + \frac{1}{\sqrt{2}}\right)^{200}.

Find: The ordered pair (ℓ,n)(\ell, n) such that

200C99Ka=2ℓmn\frac{^{200}C_{99}K}{a} = \frac{2^\ell m}{n}

where mm and nn are odd.

For (1+x)99(1+x)^{99}, the sum of coefficients of odd powers is

K=298K = 2^{98}

The middle term in the expansion of

(2+12)200\left(2 + \frac{1}{\sqrt{2}}\right)^{200}

is

T2002+1=200C100(2)100(12)100T_{\frac{200}{2}+1} = {}^{200}C_{100}(2)^{100}\left(\frac{1}{\sqrt{2}}\right)^{100}

So,

a=200C100⋅250a = {}^{200}C_{100} \cdot 2^{50}

Now,

200C99Ka=200C99⋅298200C100⋅250=200C99200C100⋅248\frac{{}^{200}C_{99}K}{a} = \frac{{}^{200}C_{99} \cdot 2^{98}}{{}^{200}C_{100} \cdot 2^{50}} = \frac{{}^{200}C_{99}}{{}^{200}C_{100}} \cdot 2^{48}

Using

200C99200C100=100101\frac{{}^{200}C_{99}}{{}^{200}C_{100}} = \frac{100}{101}

we get

200C99Ka=100101⋅248\frac{{}^{200}C_{99}K}{a} = \frac{100}{101} \cdot 2^{48}

Since

100=4⋅25=22⋅25100 = 4 \cdot 25 = 2^2 \cdot 25

therefore

100101⋅248=25101⋅250\frac{100}{101} \cdot 2^{48} = \frac{25}{101} \cdot 2^{50}

Here 2525 and 101101 are odd. Hence,

ℓ=50,n=101\ell = 50, \quad n = 101

Therefore, the ordered pair is (50,101)(50, 101) and the correct option is C.

The solution states "The Correct Option is B", but the worked steps lead to (50,101)(50, 101), which matches option C. By the solution working, C is the defensible answer.

Why the odd-coefficient sum is half of the total

Using binomial identities,

(1+1)99=∑r=09999Cr=299(1+1)^{99} = \sum_{r=0}^{99} {}^{99}C_r = 2^{99}

and

(1−1)99=∑r=09999Cr(−1)r=0(1-1)^{99} = \sum_{r=0}^{99} {}^{99}C_r (-1)^r = 0

So the sum of even-power coefficients equals the sum of odd-power coefficients. Therefore each is

2992=298\frac{2^{99}}{2} = 2^{98}

Hence K=298K = 2^{98}.

Common mistakes

  • Taking the middle term as T100T_{100} instead of T101T_{101}. For an expansion of the form (a+b)n(a+b)^n with even n=200n=200, the single middle term is Tn2+1=T101T_{\frac{n}{2}+1} = T_{101}.

  • Using K=299K = 2^{99} instead of 2982^{98}. The sum of odd-power coefficients is not the total sum of coefficients; it is half of the total because even and odd coefficient sums are equal for (1+x)99(1+x)^{99}.

  • Not simplifying the ratio 200C99200C100\frac{{}^{200}C_{99}}{{}^{200}C_{100}} correctly. Consecutive binomial coefficients satisfy nCrnCr+1=r+1n−r\frac{{}^{n}C_r}{{}^{n}C_{r+1}} = \frac{r+1}{n-r}, giving 100101\frac{100}{101} here.

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