I-n-2n-2n-1-I-n-1-I-0-1-Show-that-I-n-4-n-n-2-2n-1-

Question Number 171441 by alcohol last updated on 15/Jun/22

I_{n} = - \frac{2 n}{2 n + 1} I_{n - 1}

I_{0} = 1

S h o w t h a t I_{n} = \frac{{(- 4)}^{n} {(n!)}^{2}}{(2 n + 1)!}

Commented by infinityaction last updated on 15/Jun/22

I_{1} = - \frac{2}{3}, I_{2} = \frac{- 4}{5} \times \frac{- 2}{3}

I_{3} = \frac{- 6}{7} \times \frac{- 4}{5} \times \frac{- 2}{3}

p a t t e r n

I_{n} = \frac{- 2}{3} \times \frac{- 4}{5} \times \frac{- 6}{7} \dots \dots . . \frac{- 2 n}{2 n + 1}

I_{n} = \frac{- 2^{2}}{2 \times 3} \times \frac{- 4^{2}}{4 \times 5} \times \frac{- 6^{2}}{6 \times 7} \dots \dots \frac{- {(2 n)}^{2}}{2 n \times (2 n + 1)}

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

I_{n} = \frac{{(- 4)}^{n} {(n!)}^{2}}{(2 n + 1)!}