find-the-value-of-C-n-0-2-C-n-1-2-C-n-2-2-C-n-n-2-

Question Number 26133 by abdo imad last updated on 21/Dec/17

f i n d t h e v a l u e o f {(C_{n}^{0})}^{2} + {(C_{n}^{1})}^{2} + {(C_{n}^{2})}^{2} + \dots {(C_{n}^{n})}^{2} .

Commented by abdo imad last updated on 21/Dec/17

l e t p u t A_{n} = {(C_{n}^{0})}^{2} + {(C_{n}^{1})}^{2} + \dots + {(C_{n}^{n})}^{2}

w e h a v e {(1 + x)}^{2 n} = \sum_{k = 0}^{k = 2 n} x^{k} = C_{2 n}^{0} + C_{2 n}^{1} x + C_{2 n}^{2} x^{2} + \dots C_{2 n}^{2 n} x^{2 n}

b u t {(1 + x)}^{2 n} = {(1 + x)}^{n} . {(1 + x)}^{n} = (\sum_{k = 0}^{k = n} C_{n}^{k} x^{k}) (\sum_{k = 0}^{k = n} C_{n}^{k} x^{k})

= \sum_{k = 0}^{k = 2 n} λ_{k} x^{k} a n d λ_{k} = \sum_{i + j = k} a_{i} b_{j}

= \sum_{i = 0}^{i = k} a_{i} b_{k - i} = \sum_{i = 0}^{i = k} C_{n}^{i} C_{n}^{k - i}

- - > λ_{n} = \sum_{i = 0}^{i = n} C_{n}^{i} C_{n}^{n - i} = \sum_{i = 0}^{i = n} {(C_{n}^{i})}^{2}

b u t λ_{n} i s t h e c o e f f i c i e n t o f x^{n} - - > λ_{n} = C_{2 n}^{n}

A_{n} = C_{2 n}^{n} = \frac{(2 n)!}{{(n!)}^{2}} .

Answered by Tinkutara last updated on 21/Dec/17

Commented by abdo imad last updated on 21/Dec/17

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