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JEE Advanced 2022 Paper 1 · Mathematics · Binomial Theorem · medium

For the binomial coefficients with n=10, identify all the correct evaluations among the following.

One or more options correct

For the binomial coefficients with n=10n=10, identify all the correct evaluations among the following.
  1. A. ∑k=010(−1)k(10k)k+1=111\displaystyle\sum_{k=0}^{10}\frac{(-1)^k\binom{10}{k}}{k+1}=\frac1{11}
  2. B. ∑k=010(10k)k+1=21111\displaystyle\sum_{k=0}^{10}\frac{\binom{10}{k}}{k+1}=\frac{2^{11}}{11}
  3. C. ∑k=010k2(10k)=28160\displaystyle\sum_{k=0}^{10}k^2\binom{10}{k}=28160
  4. D. ∑k=08(10k)(10k+2)=(208)\displaystyle\sum_{k=0}^{8}\binom{10}{k}\binom{10}{k+2}=\binom{20}{8}
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Answer: A, C, D

(A) ∫01(1−x)10dx=∑(−1)k(10k)1k+1=111\int_0^1(1-x)^{10}dx=\sum(-1)^k\binom{10}k\frac1{k+1}=\frac1{11}: true. (B) ∫01(1+x)10dx=211−111\int_0^1(1+x)^{10}dx=\frac{2^{11}-1}{11}, so the sum is 204711≠204811\frac{2047}{11}\ne\frac{2048}{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) (10k+2)=(108−k)\binom{10}{k+2}=\binom{10}{8-k}, so by Vandermonde the sum is (208)=125970\binom{20}{8}=125970: true.
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