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Given the following two series:

$$1^3 + 2^3 + ... + n^3$$ $$0^3 + 1^3 + .... + (n-1)^3$$

I take the difference vertically of the two:

$$\left(1^3-0^3\right) + \left(2^3-1^3\right) + .... + \left(n^3-(n-1)^3\right)$$

This equals to $n^3$

If I now express this in sum notation:

$$\sum_{i=1}^n\left(i^3-(i-1)^3\right) = n^3$$

If I expand: $(i-1)^3 = i^3 - 3i^2 + 3i - 1$

Thus

$$\left(i^3 - (i-1)^3\right) = 3i^2 - 3i +1$$

My sum is now:

$$3\sum_{i=1}^n i^2 -3 \sum_{i=1}^n i + n = n^3$$ $$\sum_{i=1}^n i^2 = \frac{1}{3} \left(n^3 + 3 \frac{n(n-1)}{2} -n\right)$$

And the expression on the RHS above is not $\frac{1}{6} n (n+1) (2n+1)$

I don't want to solve the above using forward difference, I want to keep it backward.

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    $\begingroup$ $\sum_{i=1}^ni=\frac{n(n\color{red}{+}1)}2$. $\endgroup$ Commented Mar 19, 2016 at 22:32
  • $\begingroup$ I came here expecting "[s]eries of squares". $\endgroup$
    – geometrian
    Commented Mar 20, 2016 at 4:49

3 Answers 3

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$$\sum_{i=1}^n i={n(n+1)\over2}\quad\text{ not }\quad {n(n-1)\over2}$$

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    $\begingroup$ Oh God... why was I thinking about maximum number of unique pairs... $\endgroup$
    – nz_
    Commented Mar 19, 2016 at 22:43
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    $\begingroup$ @isquared-KeepitReal, it's an easy mistake to make. You gave a very clear presentation of your approach, which made it easy for fresh eyes to spot the error. $\endgroup$ Commented Mar 19, 2016 at 23:49
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    $\begingroup$ Thank you very much for spotting it. $\endgroup$
    – nz_
    Commented Mar 19, 2016 at 23:55
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Your error is when you evaluate $$\sum_{i=1}^n i = \frac{n(n-1)}{2}$$ The correct formula is $$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$ plugging this, everything works.

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Note that $\sum_{i=1}^n i = \frac 12n(n{\color{red}+}1)$, hence \begin{align*} \frac 13 \left(n^3 + 3\frac{n(n+1)}2 - n\right) &= \frac 16 n(2n^2 + 3n + 3 - 2)\\ &= \frac 16 n(2n^2 + 3n + 1)\\ &= \frac 16 n(2n+1)(n+1) \end{align*}

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