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

[Amer]

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]

One method is to use telescoping.

$$(n+1)^3-1^3$$​

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

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

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

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

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

 

[Amer]

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

nice method thanks

 

[chisigma]

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 $\chi$ $\sigma$