I'm trying to understand how to calculate $2^2 + 4^2 + \dots + n^2$. I've only succeed to upper bound it by $\dfrac {n^3} 2$. My goal is to say that it is $\Theta (n^3)$.
Thank you
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1What is 2^2+4^2+…+17^2 ? – Did Mar 03 '16 at 15:21
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3Use this MSE question. – Dietrich Burde Mar 03 '16 at 15:22
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Perhaps it's really a duplicate, but I don't understand the general term of the summation: why is $3^2$ missing? Is he summing over just the even numbers? If so, why does the sum end with $n^2$ ? – Mar 04 '16 at 02:31
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$\textbf{hint}$ your series is $$ \sum_{k=1}^n(2k)^2 = 4\sum_{k=1}^nk^2 = ? $$ (Assuming that its squaring the even numbers - add more terms so we can see what the true series is)
Chinny84
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Im guessing you meant the sigma to n/2 right ? I've did the calculation and i got it is ((n³∕2)+n²+(n²/2)+n))/3. i guess now i can say the limit is n^3 – limitless Mar 03 '16 at 15:31
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ah, so that $n^2$ makes sense now. But yes you can sum up if $n$ is even. – Chinny84 Mar 03 '16 at 15:35