Find n?

Calculus Level 3

If ∑ r = 0 n ( r 2 + r + 1 ) r ! = 2007 × 2007 ! \displaystyle \sum_{r=0}^n (r^{2} +r +1)r! = 2007×2007! , find n n ?

None 2006 143 2008 2005

Gaurav Raj by Gaurav Raj , 28, India

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1 solution

S = ∑ r = 0 n ( r 2 + r + 1 ) r ! = ∑ r = 0 n ( r 2 + 2 r + 1 − r ) r ! = ∑ r = 0 n ( ( r + 1 ) 2 − r ) r ! = ∑ r = 0 n ( ( r + 1 ) ( r + 1 ) ! − r r ! ) = ( n + 1 ) ( n + 1 ) ! \begin{aligned} S & = \sum_{r=0}^n \left(r^2+r+1\right)r! \\ & = \sum_{r=0}^n \left(r^2+2r+1-r\right)r! \\ & = \sum_{r=0}^n \left((r+1)^2-r\right)r! \\ & = \sum_{r=0}^n \left((r+1)(r+1)!-rr!\right) \\ & = (n+1)(n+1)! \end{aligned}

Therefore,

( n + 1 ) ( n + 1 ) ! = 2007 × 2007 ! ⟹ n = 2006 \begin{aligned} (n+1)(n+1)! & = 2007 \times 2007! \\ \implies n & = \boxed{2006} \end{aligned}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

@Chew-Seong Cheong , Limit should be *r=0 and not from r=1

Gaurav Raj - 3 years, 3 months ago

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Thanks. I have changed it.

Chew-Seong Cheong - 3 years, 3 months ago

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