Suppose $\operatorname{det}\left[\begin{array}{cc}\sum_{k=0}^{n} k & \sum_{k=0}^{n}{ }^{n} C_{k} k^{2} \\ \sum_{k=0}^{n}{ }^{n} C_{k} k & \sum_{k=0}^{n}{ }^{n} C_{k} 3^{k}\end{array}\right]=0$, holds for some positive integer $n$. Then $\sum_{k=0}^{n} \frac{{ }^{n} C_{k}}{k+1}$ equals

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Binomial Theorem

Suppose

det [\sum_{k = 0}^{n} k \sum_{k = 0}^{n}^{n} C_{k} k \sum_{k = 0}^{n}^{n} C_{k} k^{2} \sum_{k = 0}^{n}^{n} C_{k} 3^{k}] = 0

, holds for some positive integer

n

. Then

\sum_{k = 0}^{n} \frac{^{n} C _{k}}{k + 1}

equals

Answer: 6.2

Solution

Using standard binomial sum formulas:

\sum_{k = 0}^{n} k = \frac{n ( n + 1 )}{2}

\sum_{k = 0}^{n}^{n} C_{k} k^{2} = n (n - 1) 2^{n - 2} + n 2^{n - 1}

\sum_{k = 0}^{n}^{n} C_{k} k = n 2^{n - 1}

\sum_{k = 0}^{n}^{n} C_{k} 3^{k} = 4^{n}

. The determinant equation gives

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

. Simplifying:

n^{2} - 3 n - 4 = 0

, so

n = 4

. Then

\sum_{k = 0}^{4} \frac{^{4} C _{k}}{k + 1} = \frac{1}{5} \sum_{k = 0}^{4}^{5} C_{k + 1} = \frac{1}{5} (2^{5} - 1) = \frac{31}{5} = 6.20

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Numerical answer

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