if: U_0 =0 , U_(n+1) = (2n+2)U_n +2n+1 find U


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Question Number 168585 by mkam last updated on 13/Apr/22

$i f : U_{0} = 0, U_{n + 1} = (2 n + 2) U_{n} + 2 n + 1 f i n d U_{n} ?$
Commented by mr W last updated on 13/Apr/22

$U_{n} = 2^{n} n! - 1$
Commented by mkam last updated on 13/Apr/22

$c a n y o u g i v e m e s t e p b y s t e p s i r$
	Answered by mr W last updated on 13/Apr/22

	$U_{n + 1} + 1 = (2 n + 2) U_{n} + 2 n + 2$ $U_{n + 1} + 1 = 2 (n + 1) (U_{n} + 1)$ $\frac{U_{n + 1} + 1}{U_{n} + 1} = 2 (n + 1)$ $\frac{U_{n} + 1}{U_{n - 1} + 1} = 2 n$ $\frac{U_{n - 1} + 1}{U_{n - 2} + 1} = 2 (n - 1)$ $. . . . . .$ $\frac{U_{2} + 1}{U_{1} + 1} = 2 \times 2$ $\frac{U_{1} + 1}{U_{0} + 1} = 2 \times 1$ $\Rightarrow \frac{U_{n} + 1}{U_{n - 1} + 1} \times \frac{U_{n - 1} + 1}{U_{n - 2} + 1} \times . . . \times \frac{U_{1} + 1}{U_{0} + 1} = 2^{n} n \times (n - 1) \times . . . 2 \times 1$ $\Rightarrow \frac{U_{n} + 1}{U_{0} + 1} = 2^{n} n!$ $U_{n} = (U_{0} + 1) 2^{n} n! - 1$ $\Rightarrow U_{n} = 2^{n} n! - 1$
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