Mathematical Induction

Quote:

Originally Posted by stakr

Hi everyone..i am struggling with this question..Any help would be appreciated

Prove by mathematical induction that

http://img220.imageshack.us/img220/1937/formula.png

[img=http://img220.imageshack.us/img220/1937/formula.th.png]

First show that the basis holds:

$P\!\left(1\right)\,:\,\sum_{k=1}^1\left(3k^2+3k+1\ right)=7=8-1=\left(1+1\right)^3-1$

Now assume that $\sum_{k=1}^n\left(3k^2+3k+1\right)=\left(n+1\right )^3-1$ is true for some $n\in\mathbb{N}$

Now,
$\begin{aligned}\sum_{k=1}^{n+1}\left(3k^2+3k+1\rig ht)&=\sum_{k=1}^n\left(3k^2+3k+1\right)+3\left(n+1 \right)^2+3\left(n+1\right)+1\\ &= \left(n+1\right)^3-1+3\left(n+1\right)^2+3\left(n+1\right)+1\\&=n^3+6 n^2+12n+8-1\\&=\left(n^32^0+3\cdot n^22^1+3\cdot n^12^2+n^02^3\right)-1\\&=\left(n+2\right)^3-1\end{aligned}$

This completes the induction step.

(The second to last line is a result of the binomial theorem for cubes: $(x+y)^3=x^3y^0+3x^2y^1+3x^1y^2+x^0y^3$ )

Does this make sense?