Sum it very quickly

Algebra Level 4

$\large \sum _{n=1}^{2017} \dfrac {n^2 + n - 1}{(n+2)!} = \dfrac 12 - \dfrac a{b!}$

The equation above holds true for some coprime positive integers $a$ and $b$ . Find $a + b$ .

Notation: $!$ is the factorial notation. For example, $8! = 1\times2\times3\times\cdots\times8$ .

1 solution

Rishabh Jain
Nov 28, 2016

The sum is : $\large\sum_{n=1}^{2017}\dfrac{\overbrace{\underbrace{n^2+n-1}}^{ n(n+2)-(n+1)}}{(n+2)!}$

$= \sum_{n=1}^{2017}\left\{\dfrac{n(n+2)}{(n+2)!}-\dfrac{(n+1)}{(n+2)!}\right\}$

$= \sum_{n=1}^{2017}\left\{\dfrac{n}{(n+1)!}-\dfrac{(n+1)}{(n+2)!}\right\}$

$\small{\color{#0C6AC7}{\text{(A Telescopic Series)}}}$

$=\dfrac 12-\dfrac{2018}{2019!}$

$\therefore 2018+2019=\boxed{4037}$

tHANKS a lot.

i WAS unable to break it into difference of 2 terms. But you did it so quickly.

Priyanshu Mishra - 4 years, 6 months ago

@Rishabh Cool , help me in this one also

https://brilliant.org/problems/another-big-blow-summation/

Priyanshu Mishra - 4 years, 6 months ago

Same solution, vote up +1

Jason Chrysoprase - 4 years, 6 months ago