Wait, factorials in Sigma? Is it even possible?

Algebra Level 3

Find the value of the below expression.

300 lim ⁡ n → ∞ ∑ k = 1 n k 3 + 5 k 2 + 7 k + 2 ( k + 3 ) ! 300\lim_{n\rightarrow\infty} \sum_{k=1}^{n} \frac{k^3+5k^2+7k+2}{(k+3)!}


The answer is 350.

Boi (보이) by Boi (보이) , 19, South Korea

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1 solution

Rajen Kapur
Jun 11, 2017

Re-write the numerator as ( k + 3 ) ( k + 2 ) ( k + 1 ) − ( k + 3 ) ( k + 2 ) + ( k + 3 ) − 1 (k+3)(k+2)(k+1)-(k+3)(k+2)+(k+3)-1 to split each term as 300 [ 1 k ! − 1 ( k + 1 ) ! + 1 ( k + 2 ) ! − 1 ( k + 3 ) ! ] 300[\frac{1}{k!}-\frac{1}{(k+1)!}+\frac{1}{(k+2)!}-\frac{1}{(k+3)!}] telescopic sum of which gives, i.e. 300 ( 1 + 1 3 ! ) = 300 ( 7 6 ) = 350 300(1+\frac{1}{3!})=300(\frac{7}{6}) = 350 answer.

7 Helpful 1 Interesting 1 Brilliant 0 Confused

In reply to Ritik Agrawal: This is done sequentially. First divide by (k+3), then (k+2) and then (k+1).

Rajen Kapur - 3 years, 12 months ago

How did you know to break it like this.

ritik agrawal - 3 years, 12 months ago

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