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Complex Plane

Prove that: 2...

Question

Prove that: $\frac{(2 n + 1)!}{n!}$ = 2ⁿ [1 · 3 · 5 ... (2n − 1) (2n + 1)]

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Solution

$LHS = \frac{(2 n + 1)!}{n!} = \frac{(2 n + 1) (2 n) (2 n - 1) . . . . (4) (3) (2) (1)}{n!} = \frac{[(1) (3) (5) . . . . . . . . . (2 n - 1) (2 n + 1)] [(2) (4) (6) . . . . . . . . . (2 n)]}{n!} = \frac{2^{n} [(1) (3) (5) . . . . . . . . . (2 n - 1) (2 n + 1)] [(1) (2) (3) . . . . . . . . . (n)]}{n!} = \frac{2^{n} [(1) (3) (5) . . . . . . . . . (2 n - 1) (2 n + 1)] [n!]}{n!} = 2^{n} [(1) (3) (5) . . . . . . . . . (2 n - 1) (2 n + 1)] = RHS Hence, p roved .$

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Q. Prove that:(2n)!/n! = { 1*3*5....(2n-1)} 2ⁿ

\frac{(2 n)!}{n} = {1 \times 3 \times 5 . . . . (2 n - 1)} 2^{n}

\frac{(2 n + 1)!}{n!} = 2^{n} {1 \cdot 3 \cdot 5... (2 n - 1) (2 n + 1)}

\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots \dots + \frac{1}{(2 n - 1) (2 n + 1)} = \frac{n}{2 n + 1}

\frac{(2 n + 1)!}{n!} = 2^{n} 1.3.5... (2 n - 1) (2 n + 1)

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