# Summation formula

Printable View

• Feb 25th 2011, 02:23 PM
jstarks44444
Summation formula
Some help with the following would be appreciated!

Find (and prove) a formula for Σ from j=1 to k of j^3

Known propositions:

Σ from j=1 to k of j = k(k+1) / 2
Σ from j=1 to k of j^2 = k(k+1)(2k+1) / 6

Thank you!
• Feb 25th 2011, 04:16 PM
Bruno J.
Note that

$\displaystyle\sum_{j=1}^{k+1}j^4 = \sum_{j=0}^{k} (j+1)^4 = \sum_{j=0}^{k}(j^4+4j^3+6j^2+4j+1)$.

Hence

$\displaystyle(k+1)^4 = 4\sum_{j=0}^kj^3 + 6\sum_{j=0}^kj^2 + 4\sum_{j=0}^kj + k+1$.

Now use what you know!
• Feb 25th 2011, 06:44 PM
Soroban
Helolo, jstarks44444!

Quote:

$\displaystyle \text{Find (and prove) a formula for: }\: \sum^n_{k=1}k^3$

$\text{Known formulas:}$

. . $\displaystyle \sum^n_{k=1}1 \:=\:n \qquad \sum^n_{k=1}k \:=\:\frac{n(n\!+\!1)}{2} \qquad \sum^n_{k=1}k^2 \:=\:\frac{k(k\!+\!1)(2k\!+\!1)}{6}$

We note that: . $k^4 - (k-1)^4 \;=\;4k^3 - 6k^2 + 4k - 1$

Let $k \:=\:n,\:n\!-\!1,\:n\!-\!2\: \hdots \:3,\:2,\:1$

And we have this "stack" of equations:

. . $\begin{array}{cccccccccccc}
n^4 &-& (n\!-\!1)^4 &=& 4n^3 &-& 6n^2 &+& 4n &-& 1 \\
(n\!-\!1)^4 &-& (n\!-\!2)^4 &=& 4(n\!-\!1)^3 &-& 6(n\!-\!1)^2 &+& 4(n\!-\!1) &-& 1 \\
(n\!-\!2)^4 &-& (n\!-\!3)^4 &=& 4(n\!-\!2)^3 &-& 6(n\!-\!2)^2 &+& 4(n\!-\!2) &-& 1 \\
\vdots && \vdots && \vdots && \vdots && \vdots && \vdots \\
3^4 &-& 2^4 &=& 4\!\cdot\!3^3 &-& 6\!\cdot\!3^2 &+& 4\!\cdot\!3 &-& 1 \\
2^4 &-& 1^4 &=& 4\!\cdot\!2^3 &-& 6\!\cdot\!2^2 &+& 4\!\cdot\!2 &-& 1 \\
1^4 &-& 0^4 &=& 4\!\cdot\!1^3 &-& 6\!\cdot\!1^2 &+& 4\!\cdot\!1 &-& 1 \end{array}$

Add the entire stack of equations and we have:

. . $\displaystyle n^4 \;=\;4\sum^n_{k=1} k^3 - 6\sum^n_{k=1} k^2 + 4\sum^n_{k=1} k - \sum^n_{k=1} 1$

Substitute the known formulas:

. . $\displaystyle n^4 \;=\;4\sum k^3 - 6\!\cdot\!\frac{n(n+1)(2n+1)}{6} + 4\!\cdot\!\frac{n(n+1)}{2} - n$

. . $\displaystyle n^4 \;=\;4\sum k^3 - n(n+1)(2n+1) + 2n(n+1) - n$

Then we have:

. . $\displaystyle 4\sum k^3 \;=\;n^4 + n + n(n+1)(2n+1) - 2n(n+1)$

. . . . . . . . $=\;n(n^3+1) + n(n+1)(2n+1) - 2n(n+1)$

. . . . . . . . $=\;n(n+1)(n^2-n+1) + n(n+1)(2n+1) - 2n(n+1)$

. . . . . . . . $=\; n(n+1)\bigg[(n^2-n+1) + (2n+1) - 2 \bigg]$

. . . . . . . . $=\;n(n+1)(n^2+n)$

. . . . . . . . $=\;n^2(n+1)^2$

$\displaystyle \text{Therefore: }\;\sum^n_{k=1} k^3 \;=\;\frac{n^2(n+1)^2}{4}$

• Feb 25th 2011, 09:19 PM
chisigma
Quote:

Originally Posted by jstarks44444
Some help with the following would be appreciated!

Find (and prove) a formula for Σ from j=1 to k of j^3

Known propositions:

Σ from j=1 to k of j = k(k+1) / 2
Σ from j=1 to k of j^2 = k(k+1)(2k+1) / 6

Thank you!

Let's write...

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

... where $n \in \mathbb{N}$ and $k \in \mathbb{N}$. Is...

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

Now if You know $S_{1,n}$ and $S_{2,n}$ the (2) permits You to find $S_{3,n}$ ... and after that $S_{4,n}$ , $S_{5,n}$ and so one (Wink)...

Kind regards

$\chi$ $\sigma$