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JEE Advanced 2019 Paper 2 · Mathematics · Binomial Theorem · medium

Consider the sums U=Σk=0^10(binom10k)/(k+1), V=Σk=0^10((-1)^kbinom10k)/(k+1), W=Σk=0^10kbinom10k and Z=Σk=0^10binom10k^2.

One or more options correct

Consider the sums U=∑k=010(10k)k+1U=\sum_{k=0}^{10}\dfrac{\binom{10}{k}}{k+1}, V=∑k=010(−1)k(10k)k+1V=\sum_{k=0}^{10}\dfrac{(-1)^k\binom{10}{k}}{k+1}, W=∑k=010k(10k)W=\sum_{k=0}^{10}k\binom{10}{k} and Z=∑k=010(10k)2Z=\sum_{k=0}^{10}\binom{10}{k}^2. Which of the following statements is (are) correct?
  1. A. U=204711U=\dfrac{2047}{11}
  2. B. V=0V=0
  3. C. W=10240W=10240
  4. D. Z=(2010)Z=\dbinom{20}{10}
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Answer: A, D

Use 1k+1(10k)=111(11k+1)\frac{1}{k+1}\binom{10}{k}=\frac{1}{11}\binom{11}{k+1}. (A) Then U=111∑k=010(11k+1)=211−111=204711U=\frac1{11}\sum_{k=0}^{10}\binom{11}{k+1}=\frac{2^{11}-1}{11}=\frac{2047}{11}: correct. (B) Similarly V=111∑k=010(−1)k(11k+1)=111[1−∑j=011(−1)j(11j)]=111(1−0)=111V=\frac1{11}\sum_{k=0}^{10}(-1)^k\binom{11}{k+1}=\frac1{11}\left[1-\sum_{j=0}^{11}(-1)^j\binom{11}{j}\right]=\frac1{11}(1-0)=\frac1{11}, which is not 00: incorrect. (Equivalently V=∫01(1−x)10dxV=\int_0^1(1-x)^{10}dx.) (C) W=10⋅29=5120W=10\cdot2^{9}=5120, since k(10k)=10(9k−1)k\binom{10}{k}=10\binom9{k-1}; the value 1024010240 doubles the correct one: incorrect. (D) By Vandermonde's identity, ∑k(10k)(1010−k)=(2010)\sum_k\binom{10}{k}\binom{10}{10-k}=\binom{20}{10}, and (10k)=(1010−k)\binom{10}{k}=\binom{10}{10-k}, so Z=(2010)=184756Z=\binom{20}{10}=184756: correct. Answer: A and D.
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Original question written in the JEE Advanced 2019 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