Question

Series help.

Answer 1

\[\Large F(n) = 1*(1!+n)+2*(2!+n) +...n*(n!+n)\]

Answer 2

I need to find the sum of this series given the value of n.

Answer 3

\[\sum\limits_{k=1}^n k(k!+n)\]

Answer 4

yep..that's the correct general term.
How to find the sum?

Answer 5

Easy!
\[\begin{align}\sum\limits_{k=1}^n k(k!+n) &=\sum\limits_{k=1}^n k*k! + \sum\limits_{k=1}^nkn\\~\\
&= \sum\limits_{k=1}^n (k+1-1)*k! + n\sum\limits_{k=1}^nk\\~\\
&= \sum\limits_{k=1}^n ((k+1)!-k!) + n\sum\limits_{k=1}^nk\\~\\

\end{align}\]

Answer 6

are you a fan of telescoping series ?

Answer 7

if so, evaluating first sum is a piece of cake..
second sum is pretty obvious - just the sum of first n natiral numbers

Answer 8

I am trying to get a direct answer :P I want to write a program to do this, so I just input the formula and it calculates it.

Answer 9

I'll give a hint for evaluating first sum : \[\sum\limits_{k=1}^n (f(k+1) - f(k)) = f(2) - f(1) + f(3) - f(2) + \cdots + f(n+1) - f(n) \\~\\= f(n+1) - f(1)\]

Answer 10

:O yeah telescoping series..terms will cancel out..and ill get like 2 terms in the end

Answer 11

yes rest is bit easy..

Answer 12

\[\begin{align}\sum\limits_{k=1}^n k(k!+n) &=\sum\limits_{k=1}^n k*k! + \sum\limits_{k=1}^nkn\\~\\
&= \sum\limits_{k=1}^n (k+1-1)*k! + n\sum\limits_{k=1}^nk\\~\\
&= \sum\limits_{k=1}^n ((k+1)!-k!) + n\sum\limits_{k=1}^nk\\~\\
&= (n+1)! - 1 + n \frac{n(n+1)}{2}
\end{align}\]

simplify if you want

Answer 13

yeah that's what I got too :D
thanks !