Suppose det begin bmatrix sum k=0 n k & sum k=0 n nC k k 2 \\ sum k=0 n nC k & sum k=0 n nC k 3 k end bmatrix = 0 holds for some positive integer n. Then sum k=0 n frac nC k k+1 equals.

Mathematics · Binomial Theorem

JEE Advanced 2019 — Paper 2 — Question 31

\text{Suppose det} \begin{bmatrix} \sum_{k=0}^n k & \sum_{k=0}^n {}^nC_k k^2 \\ \sum_{k=0}^n {}^nC_k & \sum_{k=0}^n {}^nC_k 3^k \end{bmatrix} = 0 \text{ holds for some positive integer n. Then }

\sum_{k=0}^n \frac{{}^nC_k}{k+1} \text{ equals. }

Numerical answer — enter this value.

\sum_{k=0}^{n} k & \sum_{k=0}^{n} {^nC_k} k^2 \\ \sum_{k=0}^{n} {^nC_k} k & \sum_{k=0}^{n} {^nC_k} 3^k \end{vmatrix} = 0$$ $$\Rightarrow \begin{vmatrix} \frac{n(n+1)}{2} & n(n+1)2^{n-2} \\ n 2^{n-1} & 4^n \end{vmatrix} = 0$$ $$\Rightarrow n(n+1) 2^{n-1} \begin{vmatrix} \frac{1}{2} & 2^{n-2} \\ n & 2^{n+1} \end{vmatrix} = 0$$

\Rightarrow n(n+1) 2^{n-1} \left( \frac{1}{2} 2^{n+1} - n 2^{n-2} \right) = 0

\Rightarrow n(n+1) 2^{n-1} 2^{n-3} (4 - n) = 0

\Rightarrow n(n+1) 2^{2n-4} (4 - n) = 0

\Rightarrow n = 0, -1, 4 \Rightarrow n = 4

\sum_{k=0}^{4} \frac{{^4C_k}}{k+1} = \sum_{k=0}^{4} \frac{1}{5} {^5C_{k+1}} = \frac{1}{5} (2^5 - 1) = \frac{31}{5} = 6.2

Answer key and solution verified before publishing.

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