Match the following sums with their correct values, where n is a positive integer. List - I List - II (P) ∑ k = 0 n k 2 ( k n ) (1) n + 1 2 n + 1 − 1 (Q) ∑ k = 0 n k + 1 ( k n ) (2) n + 1

Question

Match the following sums with their correct values, where $n$ is a positive integer.

List - I	List - II
(P) $\sum_{k = 0}^{n} k^{2} (k n)$	(1) $\frac{2 ^{n + 1} - 1}{n + 1}$
(Q) $\sum_{k = 0}^{n} \frac{( k n )}{k + 1}$	(2) $\frac{1}{n + 1}$
(R) $\sum_{k = 0}^{n} (- 1)^{k} \frac{( k n )}{k + 1}$	(3) $n (n + 1) 2^{n - 2}$
(S) $\sum_{k = 0}^{n} (k n) (k - 1 n)$ (where $(- 1 n) = 0$ )	(4) $(n - 1 2 n)$

TestHub · Accepted Answer

For (P): We use k ( k n ​ ) = n ( k − 1 n − 1 ​ ) . Then ∑ k = 0 n ​ k 2 ( k n ​ ) = n ∑ k = 1 n ​ k ( k − 1 n − 1 ​ ) = n ∑ k = 1 n ​ (( k − 1 ) + 1 ) ( k − 1 n − 1 ​ ) = n ( ( n − 1 ) 2 n − 2 + 2 n − 1 ) = n ( n + 1 ) 2 n − 2 . So (P)-(3). For (Q): Consider the expansion of ( 1 + x ) n = ∑ k = 0 n ​ ( k n ​ ) x k . Integrating from 0 to 1 gives ∫ 0 1 ​ ( 1 + x ) n d x = [ n + 1 ( 1 + x ) n + 1 ​ ] 0 1 ​ = n + 1 2 n + 1 − 1 ​ . This equals ∑ k = 0 n ​ ( k n ​ ) ∫ 0 1 ​ x k d x = ∑ k = 0 n ​ k + 1 ( k n ​ ) ​ . So (Q)-(1). For (R): Consider the expansion of ( 1 − x ) n = ∑ k = 0 n ​ ( − 1 ) k ( k n ​ ) x k . Integrating from 0 to 1 gives ∫ 0 1 ​ ( 1 − x ) n d x = [ − n + 1 ( 1 − x ) n + 1 ​ ] 0 1 ​ = 0 − ( − n + 1 1 ​ ) = n + 1 1 ​ . This equals ∑ k = 0 n ​ ( − 1 ) k ( k n ​ ) ∫ 0 1 ​ x k d x = ∑ k = 0 n ​ ( − 1 ) k k + 1 ( k n ​ ) ​ . So (R)-(2). For (S): The sum ∑ k = 0 n ​ ( k n ​ ) ( k − 1 n ​ ) can be rewritten as ∑ k = 0 n ​ ( k n ​ ) ( n − ( k − 1 ) n ​ ) = ∑ k = 0 n ​ ( k n ​ ) ( n − k + 1 n ​ ) . This is the coefficient of x n − 1 in the product of ( 1 + x ) n ( 1 + x ) n = ( 1 + x ) 2 n . The coefficient of x n − 1 in ( 1 + x ) 2 n is ( n − 1 2 n ​ ) . So (S)-(4).

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