Series Demystified 2

I get $\begin{aligned} S &= \sum_{n=1}^{50} \left( (n+2)(n+1)^2-n^2 \right) (n+1)! \\ &= \sum_{n=1}^{50} \left( (n+1)^2(n+2)! - n^2(n+1)! \right) \\ &= 51^2 \cdot 52! - 2 \end{aligned}$

so the answer is $51^2 = \fbox{2601}$ .

Let the summation be denoted by $S$ .

$S = \displaystyle \sum_{n=1}^{50}[(n^3 + 3n^2 + 5n + 2)(n+1)!]$

$\Rightarrow S = \displaystyle \sum_{n=1}^{50}[((n+2)(n^2 + n + 3) - 4)(n+1)!]$

$\Rightarrow S = \displaystyle \sum_{n=1}^{50}[(n+2)!(n^2 + n + 3) - 4(n+1)!]$

$\Rightarrow S = \displaystyle \sum_{n=1}^{50}[(n+2)!((n+3)(n-2) + 9) - 4(n+1)!]$

$\Rightarrow S = \displaystyle \sum_{n=1}^{50}[(n+3)!(n-2) + 9(n+2)! - 4(n+1)!]$

$\Rightarrow S = \displaystyle \sum_{n=1}^{50}[(n+3)!(n+4 - 6) + 9(n+2)! - 4(n+1)!]$

$\Rightarrow S = \displaystyle \sum_{n=1}^{50}[(n+4)! - 6(n+3)! + 9(n+2)! - 4(n+1)!]$

$\Rightarrow S = \displaystyle \sum_{n=1}^{50}[(n+4)! - (n+3)!] -5\sum_{n=1}^{50} [(n+3)! - (n+2)!] + 4\sum_{n=1}^{50} [(n+2)! - (n+1)!]$

There we have it : A Telescopic Series

$S = \displaystyle (5! - 4! + 6! - 5! + \ldots + 54! - 53!) - 5(4! - 3! + 5! - 4! + \ldots + 53! - 52!) + 4(3! - 2! + 4! - 3! + \ldots + 52! - 51!)$

$S = \displaystyle 54! - 4! - 5(53! - 3!) + 4(52! - 2!)$

$\Rightarrow S = 54! - 5(53!) + 4(52!) - 24 + 30 - 8 = 54! - 5(53!) + 4(52!) - 2$

Required is : $\dfrac{S+2}{52!}$

$\dfrac{S+2}{52!} = 54.53 - 5.53 + 4 = 53.49 + 4 = \boxed{2601}$

$\large \text{ DEMYSTIFIED!! }$

Hoping to write it into a telescoping series of $(n+1)!$ and $(n+2)!$ , we propose that:

$n^3+3n^2+5n+2 \equiv [f(n+2)](n+2) - [f(n+1)]$

where $f(x)$ is a quadratic polynomial.

Solving it gives us $f(x)=(x-1)^2$ .

Therefore,

$\quad\displaystyle\sum_{n=1}^{50} [(n^3+3n^2+5n+2)\times(n+1)!]$

$=\displaystyle\sum_{n=1}^{50} [(n+1)^2\times(n+2)\times(n+1)!] - \sum_{n=1}^{50} [n^2\times(n+1)!]$

$=\displaystyle\sum_{n=1}^{50} [(n+1)^2\times(n+2)!] - \sum_{n=1}^{50} [n^2\times(n+1)!]$

$=\displaystyle\sum_{n=2}^{51} [n^2\times(n+1)!] - \sum_{n=1}^{50} [n^2\times(n+1)!]$

$=\displaystyle\sum_{n=51}^{51} [n^2\times(n+1)!] - \sum_{n=1}^{1} [n^2\times(n+1)!]$

$=\displaystyle 51^2\times52! - 1^2\times2!$

$=\displaystyle 51^2\times52! - 2$

$\color{#20A900}{~~n^3+3n^2+5n+2=(n+4)(n+3)(n+2)-6(n+3)(n+2)+9(n+2)-4.\\ So~ when~ we ~multiply ~by ~(n+1)!,~ we ~get\\ \displaystyle \sum_{n=1}^{50}( n^3+3n^2+5n+2)*(n+1)\\ \qquad \qquad \qquad \qquad \qquad Telescopic ~Series.\\ \displaystyle = \sum_{n=1}^{50} (n+4)! - 6(n+3)! + 9(n+2)! - 4(n+1)!.}\\ \displaystyle= \sum_{n=1}^{50} \Big\{{\color{#3D99F6}{ (n+4)! - (n+3)!}} ~~~~~{\color{#E81990}{-~~~~~5(n+3)!+ 5(n+2)!}}~~~~~ {\color{#BA33D6}{+~~~~~4(n+2)!- 4(n+1)!}} \Big\}\\ \displaystyle \therefore { \color{#3D99F6}{\sum_{n=1}^{50}\Big\{(n+4)! - (n+3)! \Big\}=- \sum_{n=1}^1 (n+3)! - \sum_{n=1}^{49}(n+4)!+ \sum_{n=1}^{49}(n+4)!+\sum_{n=50}^{50}(n+4)!} }= - 4!+54! \\ \displaystyle \therefore { \color{#E81990}{\sum_{n=1}^{50}\Big\{ -(n+3)! + (n+2)! \Big\}= \sum_{n=1}^1 (n+2)! +\sum_{n=1}^{49}(n+3)!-\sum_{n=1}^{49}(n+3)!- \sum_{n=50}^{50}(n+3)!}}= 3!-53! \\ \displaystyle \therefore { \color{#BA33D6}{\sum_{n=1}^{50}\Big\{(n+2)! - (n+1)! \Big\}=- \sum_{n=1}^1(n+1)! - \sum_{n=1}^{49}(n+2)!+\sum_{n=1}^{49}(n+2)!+\sum_{n=50}^{50}(n+2)!}} =-2! +52!\\ \color{#20A900}{So~S = 54! - 4!~~~~~ -~~~~~ 5(53! - 3!)~~~~~ +~~~~~ 4(52! - 2!) \\ S= 54! - 24~~~~~-~~~~~5*53!~+30~~~~~ +~~~~~ 4*52! - 8 \\ S = 54! - 5(53!) + 4(52!) - 2.\\ \therefore~\dfrac{S+2}{52!}\\ =54*53-5*53+4-2+2 }\\ =\Large \color{#D61F06}{2601}.$

Got as (n+2)!(n^2+3+2n)-(n+1)!(n^2+4+2n) diagonal terms cancelled to get 2 n n! add subtract 1 to n To get telescoping series final result = 2603*52! -52!(2)-2!(7)+2!(6) so (s+2)/52!=2601

happy to do all series demystified!!

I have not yet studied telescopic series and i degenerated the sum. Means in place of summation 1 to 50. Calculate S for 1 and let the summation given for integers 1 to k be shown as S .k then the general term which we have to find is T.k = (S.K + 2)/(k + 2)! So find T.1 = 4 and T.2 = 9 . So we observe T.k =( k + 1)^2. This is an observation and i just wanted to share that one may solve many problems this way by observing patterns. So T.50 = ( 51)^2 = 2601

Algebraic manipulation gives the sum: $L.H.S. = \sum_{n=1}^{50} (n+4)! - 6(n+3)! + 9(n+2)! - 4(n+1)!$ $= 54! - 5. 53! + 4. 52! - 4! + 5 . 3! - 4 . 2! = 2601 . 52! - 2$