Prove that ((1^2 ∙2!+2^2 ∙3!+3^2 ∙4!+∙∙∙+n^2 (n+1)!−2)/((n+1)!)) =n^2 +n−2


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Question Number 161861 by Rasheed.Sindhi last updated on 23/Dec/21

$Prove that$ $\frac{1^{2} \cdot 2! + 2^{2} \cdot 3! + 3^{2} \cdot 4! + \cdot \cdot \cdot + n^{2} (n + 1)! - 2}{(n + 1)!}$ $= n^{2} + n - 2$
Commented by mr W last updated on 23/Dec/21

$a f t e r s o m e t r y & t r y i g o t i t!$
Commented by Rasheed.Sindhi last updated on 24/Dec/21

$S ir I^{'} ve not got the result by simplification .$ $Actually I made sequence$ $f (1), f (2), f (3), . . .$ $and my son (Faaiz Soomro) helped$ $me to get general formula for f (n) :$ $f (n) = \frac{1^{2} \cdot 2! + 2^{2} \cdot 3! + 3^{2} \cdot 4! + \cdot \cdot \cdot + n^{2} (n + 1)! - 2}{(n + 1)!} (s a y)$ $\begin{array}{ccccccccc} n & f (n) \\ 1 & 0 \\ 2 & 4 \\ 3 & 10 \\ 4 & 18 \\ 5 & 28 \\ 6 & 40 \\ \cdot \cdot \cdot & \cdot \cdot \cdot \\ n & n^{2} + n - 2 \end{array}$
Commented by Rasheed.Sindhi last updated on 24/Dec/21

$You can't use 'macro parameter character #' in math mode$
Commented by mr W last updated on 24/Dec/21

$i w a s w o n d e r i n g h o w y o u g o t t h a t$ $r e s u l t . t o b e h o n e s t, t o p r o v e t h e$ $r e s u l t i s n o t a b i g p r o b l e m, t h e r e a r e$ $m a n y m e t h o d s . b u t i^{'} m b a s i c a l l y$ $i n t e r e s t e d h o w \underset{k = 1}{\sum^{n}} k^{2} (k + 1)! c a n b e$ $s i m p l i f i e d a t a l l, i . e . i m a i n l y {d o n}^{'} t$ $w a n t j u s t t o p r o v e t h e r e s u l t, b u t h o w$ $t o o b t a i n t h e r e s u l t .$
Commented by mr W last updated on 24/Dec/21

$y o u r s o n i s o u t s t a n d i n g l y g o o d!$
Commented by Rasheed.Sindhi last updated on 24/Dec/21
$I also changed the problem in the following system: { ((a_0 =−2 )),((a_n =a_(n−1) +2n)) :} but have not yet solved.It′s solution will be certainly a_n =n^2 +n−2$
$I also changed the problem in the$ $following system :$ ${\begin{cases} a_{0} = - 2 \\ a_{n} = a_{n - 1} + 2 n \end{cases}$ $b u t h a v e n o t y e t s o l v e d . {I t}^{'} s s o l u t i o n$ $w i l l b e c e r t a i n l y$ $a_{n} = n^{2} + n - 2$
Commented by mr W last updated on 24/Dec/21

$g o o d t h i n k i n g! s h a r p o b s e r v a t i o n!$
Commented by Rasheed.Sindhi last updated on 24/Dec/21

$G rateful S ir!$
	Answered by mr W last updated on 23/Dec/21

	$k^{2} (k + 1)!$ $= (k + 1 - 1) k (k + 1)!$ $= (k + 1) k (k + 1)! - k (k - 1) k! - 2 k k!$ $= (k + 1) k (k + 1)! - k (k - 1) k! - 2 [(k + 1)! - k!]$ $\underset{k = 1}{\sum^{n}} k^{2} (k + 1)!$ $= \underset{k = 1}{\sum^{n}} (k + 1) k (k + 1)! - \underset{k = 1}{\sum^{n}} k (k - 1) k! - 2 [\underset{k = 1}{\sum^{n}} (k + 1)! - \underset{k = 1}{\sum^{n}} k!]$ $= \underset{k = 2}{\sum^{n + 1}} k (k - 1) k! - \underset{k = 1}{\sum^{n}} k (k - 1) k! - 2 [\underset{k = 2}{\sum^{n + 1}} k! - \underset{k = 1}{\sum^{n}} k!]$ $= (n + 1) n (n + 1)! - 2 [(n + 1)! - 1!]$ $= (n^{2} + n - 2) (n + 1)! + 2$ $\underset{k = 1}{\sum^{n}} k^{2} (k + 1)! - 2 = (n^{2} + n - 2) (n + 1)!$ $\frac{\underset{k = 1}{\sum^{n}} k^{2} (k + 1)! - 2}{(n + 1)!} = n^{2} + n - 2 = (n - 1) (n + 2) ✓$
	Commented by mr W last updated on 23/Dec/21

	$i t c a n a l s o b e s t a t e d a s$ $\frac{1^{2} \cdot 2! + 2^{2} \cdot 3! + 3^{2} \cdot 4! + \cdot \cdot \cdot + n^{2} (n + 1)! - 2}{(n + 2)!} = n - 1$
	Commented by Rasheed.Sindhi last updated on 24/Dec/21

	$T h a n X a L o t s i r!$
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