What is the sum of the squares of integers starting with 10 and till 48

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Find the sum of the squares of n from n=10 to n=48:

The sum of the first n squares is `S_n=(n(n+1)(2n+1))/6` .

To find the sum from n=10 top n=48 we take `S_48-S_9`

** Take `S_9` because we want to include `10^2` ; we want to discard n=1-9 since we counted them in `S_48` , but we do not want them in the final sum.**

`S_48-S_9=(48(49)(97))/6-(9(10)(19))/6=37739`

**The sum is 37739.**

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