k-0-C-n-k-2-

Question Number 141846 by ArielVyny last updated on 24/May/21

\sum_{k ⩾ 0} {(C_{n}^{k})}^{2}

Answered by mindispower last updated on 24/May/21

\sum_{k ⩾ 0} C_{n}^{k} . C_{n}^{k}

{(1 + x)}^{n} {(1 + x)}^{n} = \sum_{k ⩾ 0} \sum_{i ⩾ 0} C_{n}^{k} C_{n}^{i} x^{k + i} = \underset{j = 0}{\sum^{2 n}} C_{2 n}^{j} x^{j}

i f w e c o m p a r c o e f f i c i e n t o f / X^{n}

C_{2 n}^{n} = \sum_{k ⩾ 0} \sum_{i ⩾ 0} C_{n}^{k} C_{n}^{j}, i + j = n

= \sum_{k ⩾ 0} C_{n}^{k} C_{n}^{n - k} = C_{2 n}^{n} \Leftrightarrow \sum_{k ⩾ 0} {(C_{n}^{k})}^{2} = C_{2 n}^{n}