JEE Advanced 2019 Paper 2Question 1
Suppose $$\det\begin{bmatrix}\sum_{k=0}^{n}k&\sum_{k=0}^{n}{}^nC_k k^2\\\sum_{k=0}^{n}{}^nC_k k&\sum_{k=0}^{n}{}^nC_k 3^k\end{bmatrix}=0$$ holds for some positive integer $n$. Then $\sum_{k=0}^{n}\frac{{}^nC_k}{k+1}$ equals ____
Binomial Theorem · Properties and Sums of Binomial Coefficients
Answer$6.2$
Solution
Concept. Evaluate each entry with a standard binomial sum; the determinant fixes $n$. Formulas used. (1) $\sum k\,{}^{n}C_k=n\,2^{n-1}$, $\sum k^2\,{}^{n}C_k=n(n+1)\,2^{n-2}$, $\sum{}^{n}C_k\,3^k=4^n$(2) $\sum_{k=0}^{n}\dfrac{{}^{n}C_k}{k+1}=\dfrac{2^{n+1}-1}{n+1}$Step 1 - The entries. $\sum_{k=0}^{n}k=\frac{n(n+1)}{2}$, and the other three come from (1). Step 2 - The determinant. $$\frac{n(n+1)}{2}\cdot4^n-n(n+1)2^{n-2}\cdot n\,2^{n-1}=n(n+1)\,2^{2n-3}\,(4-n).$$ Step 3 - Solve. It is $0$ for a positive integer $n$ only when $n=4$. Step 4 - Required sum. By (2) with $n=4$: $\dfrac{2^5-1}{5}=\dfrac{31}{5}=6.2$. Answer: 6.2