Prove that the following equations hold for any natural n.
Prove that $(C_{n}^{o})^{2} + (C_{n}^{1})^{2} + . . . + (C_{n}^{n})^{2} = C_{2 n}^{n} .$

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Solution

For deriving our expression
we have to first simplyfy this
$({^{n} C}_{0})^{2} + 9^{n} C_{1})^{2} + (^{n} C_{3})^{2} + (^{n} C_{4})^{2} + . . . . . . . + ({^{n} C}_{n})^{2} = (C_{0} C_{n} + C_{2} C_{n - 1} + . . . . . . . . . + C_{n} C_{0}) (∴^{n} C_{0} =^{n} C_{n})$

$= C o f f i c i e n t o f x^{n} i n {(1 + x)}^{n} {(1 + x)}^{n} = C o f f i c i e n t o f x^{n} i n {(1 + x)}^{2 n} =^{2 n} C_{n}$

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\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . \frac{1}{n (n + 1)} = \frac{n}{n + 1} .

(1 - \frac{1}{4}) (1 - \frac{1}{9}) (1 - \frac{1}{16}) . . . (1 - \frac{1}{n^{2}}) = \frac{n + 1}{2 n} .

2^{n} > n^{2}

n > 5.

Δ A B C

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