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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} + \dots + {(\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}) + \dots + \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} + \dots + {(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} + \dots + {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} + \dots + {({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 \dots

{({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} + \dots + {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} + \dots + \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 \dots