How do you find the sum of $Sigma [(i-1)^2+(i+1)^3]$ where i is [1,4]?

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Answer 1 · 2017-11-13T17:23:53

$238$

$sum_(i=1)^4 [(i-1)^2+(i+1)^3]$

$=sum_(i=1)^4 (i^2-2i+1+i^3+3i^2+3i+1)$

$=sum_(i=1)^4 (i^3+4i^2+i+2)$

$=((4*5)/2)^2+4*(4*5*9)/6+(4*5)/2+2*4$

$=238$

How do you find the sum of Sigma [(i-1)^2+(i+1)^3]Sigma [(i-1)^2+(i+1)^3] where i is [1,4]?