# proving the summation of a finite series .

• Feb 9th 2009, 01:10 AM
proving the summation of a finite series .
FInd $\displaystyle \sum^{n}_{r=1}r^3$

My attempt :

Using this identity : $\displaystyle r^4-(r-1)^4=4r^3-6r^2+4r-1$

$\displaystyle \sum^{n}_{r=1}r^4-(r-1)^4=\sum^{n}_{r=1}4r^3-6r^2+4r-1$

The LHS can be written as $\displaystyle n^4$ .

$\displaystyle 4\sum^{n}_{r=1}r^3=n^4+n(n+1)(2n+1)+2n(n+1)-n$

but it is not anywhere nearer to $\displaystyle \frac{1}{4}n^2(n+1)^2$ ,there must be some problem with my proving .
• Feb 9th 2009, 02:23 AM
mr fantastic
Quote:

FInd $\displaystyle \sum^{n}_{r=1}r^3$

My attempt :

Using this identity : $\displaystyle r^4-(r-1)^4=4r^3-6r^2+4r-1$

$\displaystyle \sum^{n}_{r=1}r^4-(r-1)^4=\sum^{n}_{r=1}4r^3-6r^2+4r-1$

The LHS can be written as $\displaystyle n^4$ .

$\displaystyle 4\sum^{n}_{r=1}r^3=n^4+n(n+1)(2n+1)+2n(n+1)-n$

but it is not anywhere nearer to $\displaystyle \frac{1}{4}n^2(n+1)^2$ ,there must be some problem with my proving .

You have made a couple of sign errors. The last line should be:

$\displaystyle 4\sum^{n}_{r=1}r^3 = n^4+n(n+1)(2n+1) {\color{red} - } 2n(n+1) {\color{red} + } n$

and the right hand side obligingly factorises into $\displaystyle n^2 (n+1)^2$.
• Feb 10th 2009, 05:44 AM
Thanks Mr F , i try to factorise and i still can't get the proof right .

$\displaystyle 4\sum^{n}_{r=1}r^3 = n^4+n(n+1)(2n+1)-2n(n+1)+n$
• Feb 10th 2009, 09:46 AM
Soroban

I'll modify your work ... and include all the steps.

Quote:

FInd: .$\displaystyle \sum^{n}_{r=1}r^3$

Using this identity: .$\displaystyle r^4-(r-1)^4\:=\:4r^3-6r^2+4r-1$

. . we have: .$\displaystyle \sum^{n}_{r=1}\bigg[r^4-(r-1)^4\bigg]\;=\;\sum^{n}_{r=1}\bigg[4r^3-6r^2+4r-1\bigg]$

The LHS can be written as $\displaystyle n^4$ .

We have: .$\displaystyle n^4 \;=\;4\sum r^3 - 6\sum r^2 + 4\sum r - \sum 1$

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

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

$\displaystyle \text{We have: }\;4\sum r^3 \;=\;\underbrace{n^4 + n}_{\quad\searrow} + n(n+1)(2n+1) - 2n(n+1)$

. . . . . . $\displaystyle 4\sum r^3 \;=\;\overbrace{n(n+1)(n^2-n+1)} + n(n+1)(2n+1) - 2n(n+1)$

Factor: . .$\displaystyle 4\sum r^3 \;=\;n(n+1)\bigg[(n^2-n+1) + (2n+1) - 2\bigg]$

. . . . . . .$\displaystyle 4\sum r^3 \;=\;n(n+1)\bigg[n^2+n\bigg] \;=\;n(n+1)\cdot n(n+1) \;=\;n^2(n+1)^2$

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

• Feb 10th 2009, 06:08 PM
$\displaystyle n^4+n=n(n+1)(n^2-n+1)$ , is it an algebraic identity ?
Yes. Use the fact that: $\displaystyle {\color{red}a^3 + b^3} = {\color{blue}(a+b)(a^2 - ab + b^2)}$
So: $\displaystyle n^4 + n \ = \ n({\color{red}n^3 + 1}) \ = \ n {\color{blue}(n+1)(n^2 - n + 1)}$