Use the rules for sums of powers of integers to compute the sum. $$ \sum_{j=7}^{21} j^2 $$
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Step 1
We know the formula for the sum of the first $n$ squares: $$ \sum_{j=1}^{n} j^2 = \frac{n(n+1)(2n+1)}{6} $$ Step 2: The given sum starts from $j=7$. We can rewrite the sum as the difference of two sums starting from $j=1$: $$ \sum_{j=7}^{21} j^2 = \sum_{j=1}^{21} Show more…
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