Telescopic series - general term

Question

Can you please help me find a general term for:

$$1(2) + 2(3) + 3(4) +\dotsb + (n-1)(n)$$

And please give an explanation of how you have derived the general term.

Answer 1

Note that $\frac{n^3}{3}-\frac{(n-1)^3}{3}=n(n-1)+\frac{1}{3}$ as noted by Olivier Oloa.

Now, $1\cdot 2=\frac{2^3}{3}-\frac{1^3}{3}-\frac{1}{3}.$

Then, $2\cdot 3=\frac{3^3}{3}-\frac{2^3}{3}-\frac{1}{3}.$

So, $1\cdot 2+2\cdot 3=\frac{3^3}{3}-\frac{1^3}{3}-\frac{1}{3}\cdot 2.$

Therefore, $1\cdot 2+2\cdot 3+\dots +(n-1)n=\frac{n^3}{3}-\frac{1^3}{3}-(n-1)\frac{1}{3}=\boxed{\frac{n^3-n}{3}}.$

Answer 2

Hint. One may observe that, for $n=1,2,\cdots$, we have $$ \frac{n^3}3-\frac{(n-1)^3}3=n(n-1)+\frac13. $$ This comes from the fact that $$ (n+1)^3=n^3+3n(n+1)+1, $$ then one makes $n \to n-1$.

Answer 3

You can write the sequence as

$$1(1+1) + 2(2+1) + \cdots +(n-1)((n-1)+1)$$

$$= 1^2+1 +2^2+2 + \cdots +(n-1)^2+(n-1)$$

$$ = (1^2+2^2+\cdots + (n-1)^2)+(1+2+\cdots+(n-1)$$

$$= \frac{(n-1)(n)(2n-1)}{6}+\frac{(n-1)(n)}{2}$$

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

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

$$ = \frac{(n-1)n(n+1)}{3}.$$

At the 3rd step, we use standard formula for the sums of squares and first powers.