if-U-0-0-U-n-1-2n-2-U-n-2n-1-find-U-n-

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)

\dots \dots

\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 \dots \times \frac{U_{1} + 1}{U_{0} + 1} = 2^{n} n \times (n - 1) \times \dots 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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