find the next 4 term and the n^(th) term 1,2,5,26.......


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Question Number 28743 by Cheyboy last updated on 29/Jan/18

$f i n d t h e n e x t 4 t e r m a n d t h e$ $n^{t h} t e r m$ $1, 2, 5, 26 . . . . . . .$
Commented by Tinkutara last updated on 29/Jan/18

$a_{n} = a_{n - 1}^{2} + 1$
Commented by Cheyboy last updated on 29/Jan/18

$t h a n k u s i r$
	Answered by ajfour last updated on 30/Jan/18

	$1, 2, 5, 26, 81, 186, 357, 610$ $1, 3, 21, 55, 105, 171, 253$ $2, 18, 34, 50, 66, 82$ $L e t t h e f i r s t s e r i e s 1, 2, 5, 26, . . .$ $b e T s e r i e s .$ $m i d d l e o n e S s e r i e s$ $l a s t o n e R s e r i e s$ $R_{n} = 2 + 16 (n - 1) = 16 n - 14$ $\underset{n = 1}{\sum^{n}} R_{n} = \underset{n = 1}{\sum^{n}} (S_{n + 1} - S_{n}) = \underset{n = 1}{\sum^{n}} (16 n - 14)$ $\Rightarrow S_{n + 1} - S_{1} = 8 n (n + 1) - 14 n$ $a s S_{1} = 1$ $S_{n + 1} = 8 n (n + 1) - 14 n + 1$ $S_{n + 1} = T_{n + 2} - T_{n + 1}$ $\Rightarrow \underset{n = 1}{\sum^{n}} (T_{n + 2} - T_{n + 1}) = \underset{n = 1}{\sum^{n}} [8 n (n + 1) - 14 n + 1]$ $= T_{n + 2} - T_{2} = \frac{8 n (n + 1) (n + 2)}{3} - 7 n (n + 1) + n$ $a s T_{2} = 2$ $T_{n + 2} = \frac{8 n (n + 1) (n + 2)}{3} - 7 n (n + 1)$ $+ n + 2$ $T_{n} = \frac{8 n (n - 1) (n - 2)}{3}$ $- 7 (n - 1) (n - 2) + n$ $= \frac{(n - 1) (n - 2) (8 n - 21)}{3} + n .$
	Commented by Cheyboy last updated on 29/Jan/18

	$t h a n k a l o t s i r$
	Commented by Cheyboy last updated on 29/Jan/18

	$s i r w h a t i s u r n^{t h}$
	Commented by Cheyboy last updated on 30/Jan/18

	$t h a n k u v r y m u c h s i r$
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