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 √2)^200.…
JEE Main 2023 — Mathematics Algebra
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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+21)200. If aC99200K=n2lm, where m and n are odd numbers, then the ordered pair (l,n) is equal to:
Official previous-year question
Held on 29 Jan 2023 · Verified 6 Jul 2026.
Options
A
(50,51)
B
(51,99)
C
(50,101)
D
(51,101)
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Solution
In the binomial expansion of
(1+x)99, the sum of odd coefficients of x is given by
K=C1+C3+….+C99=2299−0=298
Also, a=Middle term in the binomial expansion of (2+21)200 is
T2200+1=T100+1=C100200(2)100(21)100
⇒T100+1=C100200⋅250
So,
aC99200K
=C100200×250C99200×298=101100×248
=10125×250=(nm)2l
On comparing, we get
l=50,m=25andn=101
Therefore, (l,n)≡(50,101).
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