2

Evaluate the sum:

$$\sum_{i=1}^{25} 2i(i-1)$$

All I could do is:

$$2 \sum_{i=1}^{25} i (i-1)$$

What can I do after this?

Is there a way to evaluate without inserting every single integers?

Thank you

GFauxPas
  • 5,047
didgocks
  • 1,239
  • 5
    $\displaystyle\sum_{i=1}^{n}i^2=\frac16n(n+1)(2n+1)$, $\displaystyle\sum_{i=1}^{n}i=\frac12n(n+1)$ – Kenny Lau Jun 24 '16 at 14:06
  • Did you get that by expanding: $i(i-1)$ to $i^2 - i$? – didgocks Jun 24 '16 at 14:07
  • 4
    @didgocks Yes, that is correct. Alternatively, note that $(i+1)i(i-1)-i(i-1)(i-2)=3i(i-1)$, so you have a standard collapsing sum. That is worth remembering, because it is easier to generalise than the sum of squares, cubes, etc – almagest Jun 24 '16 at 14:10

3 Answers3

3

Expanding on @Kenny Lau's comment, $$ \begin{align} 2\sum_{i=1}^{n} i(i - 1) &= 2\bigg(\sum_{i=1}^{n} i^2 - \sum_{i=1}^{n} i\bigg) \\ &= 2\bigg(\frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2}\bigg) \\ &= 2\bigg(\frac{n(n+1)(2n+1 - 3)}{6}\bigg) \\ &= \frac{n(n+1)(2n - 2)}{3} \end{align}$$

You can plug in $n = 25$ for the answer.

shardulc
  • 4,562
2

There are general formulas which can be proven by induction.

You can use $\sum_{i=1}^n i = \frac{n(n+1)}{2}$ and $\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$, which are well-known.

Stijn Cambie
  • 163
2

With a little shift in perception, you can also see this as the sum of the triangular numbers, which is the tetrahedral numbers: $$2 \sum_{i=1}^{25} i (i-1) = 4\sum_{i=1}^{24} \frac{i (i+1)}{2}$$ (note the shift in limits)

The triangular numbers are the second column of Pascal's triangle, ${n+1 \choose 2}$, and the tetrahedral numbers are the third column, ${n+2 \choose 3}$. So your sum is: $$4\cdot{26 \choose 3} = 4\cdot\frac{26\times 25\times 24}{3 \times 2 \times 1}$$

Joffan
  • 39,627