Prove-that-1-2-2-2-2-3-3-2-4-n-2-n-1-2-n-1-n-2-n-2-

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), \dots

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 :

{\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!