Find the finite sum of the square and cube exponent of integers

Amer

Active member
Hey,
it is clear for me that
$$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$$

how to find a formula for
$$\sum_{i=1}^{n} i^2$$
$$\sum_{i=1}^{n} i^3$$
Thanks

Nehushtan

New member
One method is to use telescoping.

$$\displaystyle (n+1)^3-1^3$$​

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

$$\displaystyle =\ \sum_{i=1}^n\left[3i^2+3i+1\right]$$

$$\displaystyle =\ 3\sum_{i=1}^ni^2+\frac32n(n+1)+n$$

From this you should get $$\displaystyle \sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6$$.

For $$\displaystyle \sum_{i=1}^ni^3$$, try $$\displaystyle (n+1)^4-1^4=\sum_{i=1}^n\left[(i+1)^4-i^4\right]$$.

Amer

Active member
Re: find the finite sum of the square and cube exponent of integers

nice method thanks

chisigma

Well-known member
Re: find the finite sum of the square and cube exponent of integers

Hey,
it is clear for me that
$$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$$

how to find a formula for
$$\sum_{i=1}^{n} i^2$$
$$\sum_{i=1}^{n} i^3$$
Thanks
In general if...

$\displaystyle S_{n,k}= \sum_{i=1}^{n} i^{k}$ (1)

... then...

$\displaystyle \binom{k+1}{1}\ S_{n,1} + \binom{k+1}{2}\ S_{n,2} + ... + \binom{k+1}{k}\ S_{n,k}= (n+1)^{k+1} - (n+1)$ (1)

From (1) You easily obtain...

$\displaystyle S_{n,k}= \frac{1}{\binom{k+1}{k}}\ \{(n+1)^{k+1} - (n+1) - \sum_{i=1}^{k-1}\ \binom {k+1}{i} S_{n,i}\ \}$ (2)

The (2) allows You, given $S_{n,1}$ to find $S_{n,2}$ and given $S_{n,1}$ and $S_{n,2}$ to find $S_{n,3}$ and if You want You can proceed...

Kind regards

$\chi$ $\sigma$