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JEE Advanced 2019 Paper 1 · Mathematics · Binomial Theorem · hard

Which of the following statements is (are) correct?

One or more options correct

Which of the following statements is (are) correct?
  1. A. ∑k=010k(10k)=5120\displaystyle\sum_{k=0}^{10}k\binom{10}{k}=5120
  2. B. ∑k=08(10k)(10k+2)=(208)\displaystyle\sum_{k=0}^{8}\binom{10}{k}\binom{10}{k+2}=\binom{20}{8}
  3. C. The coefficient of x4x^{4} in the expansion of (1+x+x2)6(1+x+x^{2})^{6} is 8585.
  4. D. ∑k=010(−1)kk+1(10k)=111\displaystyle\sum_{k=0}^{10}\frac{(-1)^{k}}{k+1}\binom{10}{k}=\frac{1}{11}
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Answer: A, B, D

(A) k(10k)=10(9k−1)k\binom{10}{k}=10\binom{9}{k-1}, so the sum is 10⋅29=512010\cdot2^{9}=5120: true. (B) Write (10k+2)=(108−k)\binom{10}{k+2}=\binom{10}{8-k}; then the sum is ∑k(10k)(108−k)\sum_k\binom{10}{k}\binom{10}{8-k}, the coefficient of x8x^{8} in (1+x)20(1+x)^{20} (Vandermonde), namely (208)=125970\binom{20}{8}=125970: true. (C) The coefficient of x4x^{4} in (1+x+x2)6(1+x+x^{2})^{6} is ∑j(6j)(6−j4−2j)\sum_{j}\binom{6}{j}\binom{6-j}{4-2j} over j=0,1,2j=0,1,2 (number of x2x^2 factors jj, number of xx factors 4−2j4-2j): j=0j=0: (64)=15\binom{6}{4}=15; j=1j=1: 6(52)=606\binom{5}{2}=60; j=2j=2: (62)=15\binom62=15; total 90≠8590\ne85: false. (D) 1k+1(10k)=111(11k+1)\frac{1}{k+1}\binom{10}{k}=\frac1{11}\binom{11}{k+1}, so the sum is 111∑m=111(−1)m−1(11m)=111[1−(1−1)11]=111\frac{1}{11}\sum_{m=1}^{11}(-1)^{m-1}\binom{11}{m}=\frac1{11}\left[1-(1-1)^{11}\right]=\frac1{11}: true.
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Original question written in the JEE Advanced 2019 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