$$ \sum_{k=1}^{n}k^2 $$

Looking at it geometrically, it is obvous that:

$$ \sum_{k=1}^{n}k^2 = 1(n)+3(n-1)+5(n-2)+.....+(2n-1)(n-(n-1)) $$

$$ \ =n(1+3+5+7.....+(2n-1))-(1*3+2*5+3*7+4*9....+(n-1)*(2n-3)+(n-1)*(2n-1)) $$

$$ (1+3+5+7.....+(2n-1))= n^2 $$

$$ \ \sum_{k=1}^{n}k^2=n^3-(1*3+2*5+3*7+4*9....+(n-1)*(2n-3)+(n-1)*(2n-1)) $$

Where can I go from here?

Question

$$ \sum_{k=1}^{n}k^2 $$

Looking at it geometrically, it is obvous that:

$$ \sum_{k=1}^{n}k^2 = 1(n)+3(n-1)+5(n-2)+.....+(2n-1)(n-(n-1)) $$

$$ \ =n(1+3+5+7.....+(2n-1))-(1*3+2*5+3*7+4*9....+(n-1)*(2n-3)+(n-1)*(2n-1)) $$

$$ (1+3+5+7.....+(2n-1))= n^2 $$

$$ \ \sum_{k=1}^{n}k^2=n^3-(1*3+2*5+3*7+4*9....+(n-1)*(2n-3)+(n-1)*(2n-1)) $$

Where can I go from here?

anonymous · Answer

2 unreadable lines are
=n(1+3+5+7.....+(2n-1))-(1*3+2*5+3*7+4*9....+(n-1)*(2n-3)+(n-1)*(2n-1))

n^3-(1*3+2*5+3*7+4*9....+(n-1)*(2n-3)+(n-1)*(2n-1))

experimentx · Answer

why don't you try another approach ...

anonymous · Answer

you can try recurrence !!

anonymous · Answer

$$(1*3+2*5+3*7+4*9....+(n-3)*(2n-3)+(n-1)*(2n-1))$$
Because this approach is close to finishing, bar that expression.

anonymous · Answer

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