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JEE Advanced 2017 Paper 2 · Mathematics · Binomial Theorem · hard

Consider the following sums built from binomial coefficients binom10k of the tenth row of Pascal's triangle. Which of them is/are evaluated correctly?

One or more options correct

Consider the following sums built from binomial coefficients (10k)\binom{10}{k} of the tenth row of Pascal's triangle. Which of them is/are evaluated correctly?
  1. A. ∑k=010(−1)kk+1(10k)=111\displaystyle\sum_{k=0}^{10}\frac{(-1)^k}{k+1}\binom{10}{k}=\frac1{11}
  2. B. ∑k=0101k+1(10k)=21011\displaystyle\sum_{k=0}^{10}\frac{1}{k+1}\binom{10}{k}=\frac{2^{10}}{11}
  3. C. ∑k=010k2(10k)=28160\displaystyle\sum_{k=0}^{10}k^2\binom{10}{k}=28160
  4. D. ∑k=010k(10k)2=10(198)\displaystyle\sum_{k=0}^{10}k\binom{10}{k}^2=10\binom{19}{8}
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Answer: A, C

Use 1k+1(nk)=1n+1(n+1k+1)\frac{1}{k+1}\binom{n}{k}=\frac{1}{n+1}\binom{n+1}{k+1}. (A) The alternating sum becomes 111∑k(−1)k(11k+1)=111(1−0)=111\frac1{11}\sum_{k}(-1)^k\binom{11}{k+1}=\frac1{11}\left(1-0\right)=\frac1{11}, since ∑j=011(−1)j(11j)=0\sum_{j=0}^{11}(-1)^j\binom{11}{j}=0; true. (B) The plain sum is 111(211−1)=204711\frac{1}{11}(2^{11}-1)=\frac{2047}{11}, not 210/112^{10}/11; false. (C) ∑k2(nk)=n(n+1)2n−2=10⋅11⋅256=28160\sum k^2\binom nk=n(n+1)2^{n-2}=10\cdot11\cdot256=28160; true. (D) k(nk)=n(n−1k−1)k\binom nk=n\binom{n-1}{k-1} gives ∑k(nk)2=n(2n−1n−1)=10(199)=923780\sum k\binom nk^2=n\binom{2n-1}{n-1}=10\binom{19}{9}=923780, whereas 10(198)=75582010\binom{19}{8}=755820; false. Every value was confirmed with exact Fraction arithmetic. The true pair is A with C; the identity relating (nk)/(k+1)\binom nk/(k+1) to a coefficient of the next row is the main trick.
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Original question written in the JEE Advanced 2017 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake