PrepRung
Sign inSign up free

JEE Advanced 2025 Paper 1 · Mathematics · Binomial Theorem · medium

Let n=10 and let Ck=binom10k.

One or more options correct

Let n=10n=10 and let Ck=(10k)C_k=\binom{10}{k}. Which of the following statements is/are correct?
  1. A. ∑k=010k Ck=5⋅210\displaystyle\sum_{k=0}^{10}k\,C_k=5\cdot2^{10}
  2. B. ∑k=010k2Ck=27⋅210\displaystyle\sum_{k=0}^{10}k^{2}C_k=27\cdot2^{10}
  3. C. ∑k=010(−1)kCkk+2=1132\displaystyle\sum_{k=0}^{10}\frac{(-1)^kC_k}{k+2}=\frac1{132}
  4. D. ∑k=09Ck Ck+1=(2010)\displaystyle\sum_{k=0}^{9}C_k\,C_{k+1}=\binom{20}{10}
Show answer and solution

Answer: A, C

(A) kCk=10(9k−1)kC_k=10\binom9{k-1}, so the sum is 10⋅29=5⋅21010\cdot2^9=5\cdot2^{10}: true. (B) ∑k2Ck=n(n+1)2n−2\sum k^2C_k=n(n+1)2^{n-2} (from k2=k(k−1)+kk^2=k(k-1)+k giving n(n−1)2n−2+n2n−1n(n-1)2^{n-2}+n2^{n-1}), which is 110⋅28=28160110\cdot2^8=28160, whereas 27⋅210=2764827\cdot2^{10}=27648: false. (C) Integrate (1−x)10=∑(−1)kCkxk(1-x)^{10}=\sum(-1)^kC_kx^k after multiplying by xx: ∑(−1)kCkk+2=∫01x(1−x)10dx=B(2,11)=1! 10!12!=1132\sum\dfrac{(-1)^kC_k}{k+2}=\int_0^1x(1-x)^{10}dx=B(2,11)=\dfrac{1!\,10!}{12!}=\dfrac1{132}: true. (D) Ck+1=C9−kC_{k+1}=C_{9-k}, so ∑kCkC9−k=(209)=167960\sum_{k}C_kC_{9-k}=\binom{20}{9}=167960 by Vandermonde, which is not (2010)=184756\binom{20}{10}=184756: false. Exact arithmetic confirms these values. Correct options: A and C.
Report a mistake

Original question written in the JEE Advanced 2025 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake