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The sum $1^2 + 2^2 + 3^2 + 4^2 + \cdots + n^2 = n(n+1)(2n+1) \div 6$. What is the value of $21^2 + 22^2 + \cdots + 40^2 + 41^2 + ... + 100^2$?

 Jun 3, 2023
 #1
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The value of $21^2 + 22^2 + \cdots + 40^2 + 41^2 + \cdots + 100^2$ is 208450.

 Jun 3, 2023
 #2
avatar+33604 
+1

Let \(S(n)=\frac{n(n+1)(2n+1)}{6}\)

 

Then you want  \(S(100)-S(20)\)

 Jun 3, 2023

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