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I came across a question today...

Sum to $n$ term of the series $$1^3-1.5^3+2^3-2.5^3+...$$ is?

I divided this series in two different series $$\left(1^3+2^3+3^3+...\right)-\left(1.5^3+2.5^3+3.5^3+...\right)$$

$$\Rightarrow\sum^{\frac{n+1}{2}}_{r=1}r^3-\sum^{\frac{n-1}{2}}_{r=1}\left(r+\frac{1}{2}\right)^3$$

$$\Rightarrow\sum^{\frac{n+1}{2}}_{r=1}r^3-\sum^{\frac{n-1}{2}}_{r=1}\left(r^3+\frac{1}{8}+1.5r^2+0.75r\right)$$

I solved this and couldn't get to the answer. Am I doing it in the right way?

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    $\begingroup$ Your idea is correct. Use the formulas $$\sum_{j=1}^n j=\frac{n(n+1)}{2}$$ and $$\sum_{j=1}^n j^2=\frac{n(n+1)(2n+1)}{6}$$ $\endgroup$ – Peter Apr 18 '16 at 19:31
  • $\begingroup$ @Peter Do I need to put $\dfrac{n+1}{2}$ in place of $n$ in these formulas? $\endgroup$ – manshu Apr 18 '16 at 19:33
  • $\begingroup$ Exactly that. Note, that the $r^3$-term luckily cancels out, so we do not need the formula for the cubes. $\endgroup$ – Peter Apr 18 '16 at 19:34
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Hints: You can represent the series as $$(\frac{-n}2)^3$$ Assume even number of terms $$n^3 - (n+1)^3 = -(3n^2+3n+1)$$

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