such that 1. ((n),(0) )^2 + ((n),(1) )^2 +...+ ((n),(n) )^2 =(((2n)!)/((n!)^2 )) 2. ((n),(0) )+(1/2) ((n),(1) )+...+(1/(n+1)) ((n),(n) )^2 =((2^(n+1) −1)/(n+1))


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Question Number 54741 by gunawan last updated on 10/Feb/19

$such that$ $1 . {(\begin{matrix} n \\ 0 \end{matrix})}^{2} + {(\begin{matrix} n \\ 1 \end{matrix})}^{2} + . . . + {(\begin{matrix} n \\ n \end{matrix})}^{2} = \frac{(2 n)!}{{(n!)}^{2}}$ $2 . (\begin{matrix} n \\ 0 \end{matrix}) + \frac{1}{2} (\begin{matrix} n \\ 1 \end{matrix}) + . . . + \frac{1}{n + 1} {(\begin{matrix} n \\ n \end{matrix})}^{2} = \frac{2^{n + 1} - 1}{n + 1}$
Commented by Abdo msup. last updated on 10/Feb/19

$1) w e h a v e {(1 + x)}^{2 n} = \sum_{k = 0}^{n} C_{2 n}^{k} x^{k} a n d$ ${(1 + x)}^{2 n} = {(1 + x)}^{n} {(1 + x)}^{n} = (\sum_{k = 0}^{n} C_{n}^{k} x^{k}) (\sum_{k = 0}^{n} C_{n}^{k} x^{k})$ $= \sum_{k = 0}^{2 n} c_{k} x^{2 k} w i t h c_{k} = \sum_{i + j = k} a_{i} b_{j}$ $= \sum_{i = 0}^{k} a_{i} b_{k - i} = \sum_{i = 0}^{k} C_{n}^{i} C_{n}^{k - i} = \sum_{i = 0}^{k} {(C_{n}^{i})}^{2} \Rightarrow$ $c_{n} = \sum_{i = 0}^{n} {(C_{n}^{i})}^{2} b u t c_{n} i s t h e c o e f f i c i e n t o f x^{2 n} i n t h e$ $e x p a n t i o n o f {(1 + x)}^{2 n} \Rightarrow c_{n} = C_{2 n}^{n} = \frac{(2 n)!}{{(n!)}^{2}} \Rightarrow$ ${(C_{n}^{0})}^{2} + {(C_{n}^{1})}^{2} + {(C_{n}^{3})}^{2} + . . . + {(C_{n}^{n})}^{2} = \frac{(2 n)!}{{(n!)}^{2}} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Feb/19

	$1) {(1 + x)}^{n} = {n c}_{0} + {n c}_{1} x + {n c}_{2} x^{2} + . . . + {n c}_{n} x^{n}$ ${(1 + \frac{1}{x})}^{n} = {n c}_{0} + {n c}_{1} \times \frac{1}{x} + {n c}_{2} \times \frac{1}{x^{2}} + . . + {n c}_{n} \times \frac{1}{x^{n}}$ $m u l t i p l y r i g h t h a n d s i d e a n d t a k e x i n d e p e n d e n t$ $t e r m s . .$ ${({n c}_{0})}^{2} + {({n c}_{1})}^{2} + {({n c}_{2})}^{2} + . . . + {({n c}_{n})}^{2}$ $m u l t i p l y l e f t h a n d s i d e a n d t a k e t h e c o e f f i c i e n t$ $o f x i n d e p e n t t e r m s$ ${(1 + x)}^{n} \times {(1 + \frac{1}{x})}^{n}$ $= {(1 + x)}^{n} \times {(1 + x)}^{n} \times \frac{1}{x^{n}}$ $= x^{- n} {(1 + x)}^{2 n}$ $l e t (r + 1) t h t e r m o f {(1 + x)}^{2 n} c o n t a i n s x^{n}$ $t_{r + 1} = 2 {n c}_{r} {(x)}^{r}$ $h e n c e x^{r} = x^{n}$ $r = n$ $t_{n + 1} = 2 {n c}_{n} {(x)}^{n} c o n t a i n s x^{n}$ $x^{- n} \times 2 {n c}_{n} x^{n}$ $= 2 {n c}_{n} \to x i n d e p e n d e n d e n t . . .$ ${({n c}_{0})}^{2} + {({n c}_{1})}^{2} + . . {({n c}_{n})}^{2} = 2 {n c}_{n} = \frac{(2 n)!}{n! \times n!}$ $2) \int_{0}^{1} {(1 + x)}^{n} d x = \int_{0}^{1} [{n c}_{0} + {n c}_{1} x + {n c}_{2} x^{2} + . . . + {n c}_{n} x^{n}] d x$ $∣ \frac{{(1 + x)}^{n + 1}}{n + 1} ∣_{0}^{1} =∣ {n c}_{0} x + \frac{1}{2} {n c}_{1} x^{2} + \frac{1}{3} {n c}_{2} x^{3} + . . + \frac{1}{n + 1} {n c}_{n} x^{n + 1} ∣_{0}^{1}$ $\frac{2^{n + 1} - 1}{n + 1} = {n c}_{0} + \frac{1}{2} {n c}_{1} + \frac{1}{3} {n c}_{2} + . . . + \frac{1}{n + 1} {n c}_{n}$
	Commented by gunawan last updated on 10/Feb/19

	$wow thank y o u very much Sir$
	Commented by tanmay.chaudhury50@gmail.com last updated on 10/Feb/19

	$m o s t w e l c o m e . . .$
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