U_(n+2) −2U_(n+1) +U


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Question Number 159939 by Ar Brandon last updated on 22/Nov/21

$U_{n + 2} - {2 U}_{n + 1} + U_{n} = 800$
Commented by mr W last updated on 22/Nov/21

$t w o t e r m s m u s t b e g i v e n, f o r e x a m p l e$ $U_{0}, U_{1} o r U_{1}, U_{2} .$
Commented by Ar Brandon last updated on 22/Nov/21

$Ok Sir . Let me review the question .$ $Thanks for your attention .$
	Answered by mr W last updated on 22/Nov/21

	$a s s u m e U_{0} = 1, U_{1} = 2 .$ $(U_{n + 2} - U_{n + 1}) - (U_{n + 1} - U_{n}) = 800$ $l e t V_{n} = U_{n + 1} - U_{n}$ $V_{n + 1} - V_{n} = 800$ $t h i s i s a n A . P . w i t h c o m m o n$ $d i f f e r e n c e 800 .$ $\Rightarrow V_{n} = V_{0} + 800 n$ $U_{n + 1} - U_{n} = V_{0} + 800 n$ $U_{n} - U_{n - 1} = V_{0} + 800 (n - 1)$ $. . .$ $U_{2} - U_{1} = V_{0} + 800 \times 1$ $U_{1} - U_{0} = V_{0} + 800 \times 0$ $\Rightarrow U_{n + 1} - U_{0} = (n + 1) V_{0} + 800 \times \frac{n (n + 1)}{2}$ $\Rightarrow U_{n + 1} = U_{0} + (n + 1) V_{0} + 800 \times \frac{n (n + 1)}{2}$ $o r$ $U_{n} = U_{0} + {n V}_{0} + 800 \times \frac{n (n - 1)}{2}$ $\Rightarrow U_{n} = U_{0} + n (U_{1} - U_{0}) + 400 n (n - 1)$ $w i t h U_{0} = 1, U_{1} = 2$ $\Rightarrow U_{n} = 1 + n + 400 n (n - 1)$
	Commented by Ar Brandon last updated on 23/Nov/21
	Thank you sir
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