JEE Advanced 2019 · Paper 2, Mathematics question 9

The sums.
$$\sum k=\frac{n(n+1)}2,\qquad\sum\binom nkk=n2^{n-1},\qquad\sum\binom nkk^2=n(n+1)2^{n-2},\qquad\sum\binom nk3^k=4^n.$$

The determinant.
$$\frac{n(n+1)}2\cdot4^n-n(n+1)2^{n-2}\cdot n2^{n-1}=n(n+1)4^n\left(\frac12-\frac n8\right)=0\implies n=4.$$

The required sum.
$$\sum_{k=0}^4\frac{\binom4k}{k+1}=\frac{2^5-1}5=\frac{31}5=6.2.$$
Answer: 6.2