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# Summation problem

Evaluate $$\displaystyle{\sum_{a= 3}^{6} (a-2)^2}$$ and $$\displaystyle{\sum_{a=1}^{4} a^2}$$.

     What do you notice?
Why does this work?


Note by Cedie Camomot
10 months, 1 week ago

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Well, the value of both of them is same, i.e., $$30$$.

$\displaystyle \sum^{6}_{a=3}(a-2)^2 \\ \text{Let, } n=a-2\Rightarrow a=n+2 \\ \displaystyle \sum^{n+2=6}_{n+2=3}n^2=\displaystyle \sum^{n=4}_{n=1}n^2 \text{ or } \displaystyle \sum^{4}_{n=1}n^2$ · 10 months, 1 week ago