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Compute (1^2 + 2^2 + 3^2 + ... + 10^2)/770.

 Jul 5, 2020
 #1
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sumfor(n, 1, 10, (n^2))=462 / 770 =3 / 5 

 Jul 5, 2020
 #2
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Compute (1^2 + 2^2 + 3^2 + ... + 10^2)/770.

 

In general

 

\(\Sigma_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}\)

 

When n = 10 we have: 

\(\Sigma_{k=1}^{10}k^2=\frac{10.11.21}{6}=385\)

 

And 385/770 = 1/2

 Jul 6, 2020

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