How do you find the sum of $Sigma (k+1)^2(k-3)$ where k is [2,5]?

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Answer 1 · 2017-08-18T20:28:34

$sum_(k=2)^5(k+1)^2(k-3)$

With so few terms, since we're only adding $4$ , it's easy to just plug in all $k$ values from $k=2$ to $k=5$ and add them:

$=(2+1)^2(2-3)+(3+1)^2(3-3)+(4+1)^2(4-3)+(5+1)^2(5-3)$

$=9(-1)+16(0)+25(1)+36(2)$

$=color(blue)88$

How do you find the sum of Sigma (k+1)^2(k-3)Sigma (k+1)^2(k-3) where k is [2,5]?