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- #1

- Thread starter Amer
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- Thread starter
- #1

- Mar 11, 2013

- 5

\(\displaystyle (n+1)^3-1^3\)

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

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

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

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 \(\displaystyle (n+1)^4-1^4=\sum_{i=1}^n\left[(i+1)^4-i^4\right]\).

- Thread starter
- #3

- Feb 13, 2012

- 1,704

In general if...Hey,

it is clear for me that

[tex] \sum_{i=1}^{n} i = \frac{n(n+1)}{2} [/tex]

how to find a formula for

[tex] \sum_{i=1}^{n} i^2 [/tex]

[tex] \sum_{i=1}^{n} i^3 [/tex]

Thanks

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

$\chi$ $\sigma$