How do you evaluate $20 \sum n = 1 (n + n^{3})$ $\sum _ { n = 1} ^ { 20} ( n + n ^ { 3} )$ ?

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How do you evaluate $20 \sum n = 1 (n + n^{3})$ $\sum _ { n = 1} ^ { 20} ( n + n ^ { 3} )$ ?

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Answer 1 · 2017-05-10T19:46:30

$20 \sum n = 1 (n + n^{3}) = 44310$ $sum_(n=1)^20 (n+n^3) = 44310$

Explanation:

Let:

$S = 20 \sum n = 1 (n + n^{3})$ $S = sum_(n=1)^20 (n+n^3)$

We need the standard formula:

$sum_(r=1)^n r \ = 1/2n(n+1)$
$sum_(r=1)^n r^2 = 1/4n^2(n+1)^2$

Which gives us:
$S = sum_(n=1)^20 n+sum_(n=1)^20 n^3$
$\ \ = 1/2(20)(21)+1/4(20^2)(21^2)$
$\ \ = 1/2(420)+1/4(400)(441)$
$\ \ = 210 + (100)(441)$
$\ \ = 210 + 44100$
$\ \ = 44310$

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How do you evaluate $20 \sum n = 1 (n + n^{3})$ $\sum _ { n = 1} ^ { 20} ( n + n ^ { 3} )$ ?

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