Annoying factorial

Algebra Level 3

Let s ( n ) = ∑ k = 0 n k ! ( k 2 + k + 1 ) . \large s(n)=\sum_{k=0}^{n}k!(k^2+k+1). Find the value of s ( 5 ) s(5) .


Notation: ! ! is the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 8! = 1\times2\times3\times\cdots\times8 .


The answer is 4320.

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Sergi Mlg Sabat by Sergi Mlg Sabat , 29, USA

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1 solution

Rishabh Jain
Mar 20, 2017

Write k 2 + k + 1 = ( k + 1 ) 2 − k k^2+k+1=(k+1)^2-k so that sum is ∑ k = 0 n k ! [ ( k + 1 ) 2 − k ] \sum_{k=0}^{n}k![(k+1)^2-k] = ∑ k = 0 n [ k ! ( k + 1 ) 2 − k ⋅ k ! ] =\sum_{k=0}^{n}\left[k!(k+1)^2-k\cdot k!\right] = ∑ k = 0 n [ ( k + 1 ) ⋅ ( k + 1 ) ! − k ⋅ k ! ] =\sum_{k=0}^{n}\left[(k+1)\cdot(k+1)!-k\cdot k!\right] ( T e l e s c o p i c S e r i e s ) \mathcal{(Telescopic~Series)}

= ( n + 1 ) ⋅ ( n + 1 ) ! =\boxed{(n+1)\cdot (n+1)!}

Therefore, s ( 5 ) = 6 ⋅ 6 ! = 4320 s(5)=6\cdot 6!=\boxed{4320}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Yep absolutely trivial

Ραμών Αδάλια - 4 years, 2 months ago

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