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Algebra Level 4

∑ k = 0 n \displaystyle \sum_{k=0}^n ( k 2 k^2 + k + 1 ) k! = 2007 ⋅ 2007 ! 2007 \cdot 2007!
Find n .


The answer is 2006.

Akhil Bansal by Akhil Bansal , 22, India

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1 solution

Maggie Miller
Aug 6, 2015

∑ k = 0 n ( k 2 + k + 1 ) k ! = ∑ k = 0 n ( ( k + 1 ) 2 − k ) k ! = ∑ k = 0 n ( ( k + 1 ) ⋅ ( k + 1 ) ! − k ⋅ k ! ) \displaystyle\sum_{k=0}^n(k^2+k+1)k!=\sum_{k=0}^n((k+1)^2-k)k!=\sum_{k=0}^n\left((k+1)\cdot (k+1)!-k\cdot k!\right) .

This is a telescoping series, so ∑ k = 0 n ( k 2 + k + 1 ) k ! = ( n + 1 ) ⋅ ( n + 1 ) ! − 0 ⋅ 0 ! = ( n + 1 ) ⋅ ( n + 1 ) ! \displaystyle\sum_{k=0}^n(k^2+k+1)k!=(n+1)\cdot(n+1)!-0\cdot0!=(n+1)\cdot (n+1)! . Therefore, ( n + 1 ) ⋅ ( n + 1 ) ! = 2007 ⋅ 2007 ! (n+1)\cdot(n+1)!=2007\cdot2007! , so n = 2006 n=\boxed{2006} .

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You are awsome Maggie .
Thnx for posting a solution.

Akhil Bansal - 5 years, 10 months ago

I am not getting what is telescoping series please help me .

prateek anand - 5 years, 10 months ago

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In a telescoping series, successive terms cancel.

∑ k = 0 n ( ( k + 1 ) ⋅ ( k + 1 ) ! − k ⋅ k ! ) \displaystyle\sum_{k=0}^n((k+1)\cdot(k+1)!-k\cdot k!)

= ( 1 ⋅ 1 ! − 0 ⋅ 0 ! ) + ( 2 ⋅ 2 ! − 1 ⋅ 1 ! ) + ( 3 ! ⋅ 3 ! − 2 ⋅ 2 ! ) + ⋯ =(1\cdot 1!-0\cdot 0!)+(2\cdot 2!-1\cdot 1!)+(3!\cdot3!-2\cdot2!)+\cdots

+ ( n ⋅ n ! − ( n − 1 ) ⋅ ( n − 1 ) ! ) + ( ( n + 1 ) ⋅ ( n + 1 ) ! − n ⋅ n ! ) +(n\cdot n!-(n-1)\cdot(n-1)!)+((n+1)\cdot(n+1)!-n\cdot n!)

Now rearrange the terms to put like factorials together:

− 0 ⋅ 0 ! + ( 1 ⋅ 1 ! − 1 ⋅ 1 ! ) + ( 2 ⋅ 2 ! − 2 ⋅ 2 ! ) + ⋯ -0\cdot 0!+(1\cdot 1!-1\cdot 1!)+(2\cdot 2!-2\cdot 2!)+\cdots

+ ( n ⋅ n ! − n ⋅ n ! ) + ( n + 1 ) ⋅ ( n + 1 ) ! +(n\cdot n!-n\cdot n!)+(n+1)\cdot(n+1)!

And all the middle terms cancel out, leaving − 0 ⋅ 0 ! + ( n + 1 ) ⋅ ( n + 1 ) ! -0\cdot 0!+(n+1)\cdot(n+1)! .

Maggie Miller - 5 years, 10 months ago

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I salute you thanks for guiding me

prateek anand - 5 years, 10 months ago

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