Only if this could be "Series Demystified 8"

Algebra Level 5

∑ n = 1 48 n 2 + n − 1 ( n + 2 ) ! \large \displaystyle \sum_{n=1}^{48} \dfrac{n^2 + n - 1}{(n+2)!}

If the above summation equals to S S , find 50 ! ( 1 2 − S ) 50!\left(\dfrac{1}{2} - S \right) .

Try my set .


The answer is 49.

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Vishwak Srinivasan by Vishwak Srinivasan , 24, India

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2 solutions

This problem is very similar to Sanjeet sir's Series Demystified, hence the name.

It can found that:

n 2 + n − 1 = n ( n + 1 ) − 1 = n ( n + 2 ) − ( n + 1 ) n^2 + n -1 = n(n+1) - 1 = n(n+2) - (n+1)

which is key to solving the question.

S = ∑ n = 1 48 n ( n + 2 ) − ( n + 1 ) ( n + 2 ) ! = ∑ n = 1 48 n ( n + 1 ) ! − n + 1 ( n + 2 ) ! S = \displaystyle \sum_{n=1}^{48} \dfrac{n(n+2) -(n+1)}{(n+2)!} = \sum_{n=1}^{48} \dfrac{n}{(n+1)!} - \dfrac{n+1}{(n+2)!}

This is of course a telescopic series

S = ( 1 2 ! − 2 3 ! ) + ( 2 3 ! − 3 4 ! ) + … + ( 48 49 ! − 49 50 ! ) S = \left(\dfrac{1}{2!} - \dfrac{2}{3!}\right) + \left(\dfrac{2}{3!} - \dfrac{3}{4!}\right) + \ldots + \left(\dfrac{48}{49!} - \dfrac{49}{50!} \right)

⇒ S = 1 2 ! − 49 50 ! \Rightarrow S = \dfrac{1}{2!} - \dfrac{49}{50!}

Required Answer is 50 ! ( 1 2 − S ) = 50 ! 49 50 ! = 49 50!\left(\dfrac{1}{2} - S\right) = 50!\dfrac{49}{50!} = \boxed{49}

8 Helpful 0 Interesting 0 Brilliant 0 Confused

n 2 + n − 1 = ( n + 1 ) ( n + 2 ) − 2 ( n + 2 ) + 1. ∴ ∑ n = 1 48 n 2 + n − 1 ( n + 2 ) ! = ∑ n = 1 48 { ( n + 1 ) ( n + 2 ) ( n + 2 ) ! − 2 n + 2 ( n + 2 ) ! + 1 ( n + 2 ) ! } = ∑ n = 1 48 { 1 n ! − 1 ( n + 1 ) ! + 1 ( n + 2 ) ! − 1 ( n + 1 ) ! } T h e s e a r e t w o t e l e s c o p i c s e r i e s e a c h w i t h o n e i n i t i a l t e r m , a n d a f i n a l t e r m . S = ( 1 − 1 49 ! ) + ( 1 50 ! − 1 2 ! ) ⟹ 50 ! ( 1 2 − S ) = { 1 2 − ( 1 2 − 50 + 1 ) } = 49. n^2+n-1=(n+1)(n+2)-2(n+2)+1.\\ \therefore~\displaystyle \sum_{n=1}^{48} \dfrac{n^2+n-1}{(n+2)!}=\sum_{n=1}^{48} \left\{\dfrac{(n+1)(n+2)}{(n+2)!}-2\dfrac{n+2}{(n+2)!}+\dfrac 1 {(n+2)!} \right \}\\ \displaystyle = \sum_{n=1}^{48} \left\{\dfrac 1 {n!}- \dfrac 1 {(n+1)!} + \dfrac 1 {(n+2)!} - \dfrac 1 {(n+1)!} \right \}\ \\ ~~~ These ~are ~two~ telescopic~ series~each~with~one~initial~term, ~and~a~final ~term.\\ S=\left (1-\dfrac 1 {49!} \right)~~+~~\left ( \dfrac 1 {50!} - \dfrac 1 {2!} \right) \\ ~~~~ \implies~50!(\frac 1 2-S)= \left \{\frac 1 2- (\frac 1 2- 50+ 1 ) \right\}=~\Large \color{#D61F06}{49}.

Niranjan Khanderia - 2 years, 1 month ago

8 of 8 Vishwak Srinivasan

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Yeah. Lol. Maybe you could add this as a comment. Instead of a solution.

Vishwak Srinivasan - 5 years, 11 months ago

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