I_n = −((2n)/(2n + 1)) I_(n−1) I_0 = 1 Show that I


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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
$I_(1 ) = −(2/3) , I_2 = ((−4)/5)×((−2)/3) I_3 = ((−6)/7)×((−4)/5)×((−2)/3) pattern I_n = ((−2)/3)×((−4)/5)×((−6)/7)........((−2n)/(2n+1)) I_n = ((−2^2 )/(2×3))×((−4^2 )/(4×5))×((−6^2 )/(6×7))......((−(2n)^2 )/(2n×(2n+1))) I_n = (((−1)^n 2^(2n) {1^2 ×2^2 ×3^2 ×.......n^2 })/((2n+1)!)) I_n = (((−4)^n (n!)^2 )/((2n+1)!))$
$I_{1} = - \frac{2}{3}, I_{2} = \frac{- 4}{5} \times \frac{- 2}{3}$ $I_{3} = \frac{- 6}{7} \times \frac{- 4}{5} \times \frac{- 2}{3}$ $p a t t e r n$ $I_{n} = \frac{- 2}{3} \times \frac{- 4}{5} \times \frac{- 6}{7} . . . . . . . . \frac{- 2 n}{2 n + 1}$ $I_{n} = \frac{- 2^{2}}{2 \times 3} \times \frac{- 4^{2}}{4 \times 5} \times \frac{- 6^{2}}{6 \times 7} . . . . . . \frac{- {(2 n)}^{2}}{2 n \times (2 n + 1)}$ $I_{n} = \frac{{(- 1)}^{n} 2^{2 n} {1^{2} \times 2^{2} \times 3^{2} \times . . . . . . . n^{2}}}{(2 n + 1)!}$ $I_{n} = \frac{{(- 4)}^{n} {(n!)}^{2}}{(2 n + 1)!}$
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