# Mathematical induction proof, help?

• Feb 23rd 2010, 03:26 PM
forumlurker
Mathematical induction proof, help?
S(n): 2+7+12+...+(5n+2) = 1/2(n+1)(5n+4)

S(1): LHS (5(1)+2) = 9
RHS 1/2(1+1)(5(1)+4) = 9 RHS = LHS

Assume True: S(k) = 2+7+12+...+(5k+2) = 1/2(k+1)(5k+4)

Want to Show True: S(k+1) = 2+7+12+...+(5(k+1)+2) = 1/2((k+1)+1)(5(k+1)+4)

Here's what I have so far.

LHS = 2+7+12+...+ (5k+2) + (5(k+1)+2) = 5(k+1)+ 2 + 1/2(k+1)(5k+4) by induction hypothesis

I am stuck here...if anyone can help I appreciate it.
• Feb 23rd 2010, 03:57 PM
Quote:

Originally Posted by forumlurker
S(n): 2+7+12+...+(5n+2) = 1/2(n+1)(5n+4)

S(1): LHS (5(1)+2) = 9
RHS 1/2(1+1)(5(1)+4) = 9 RHS = LHS

Assume True: S(k) = 2+7+12+...+(5k+2) = 1/2(k+1)(5k+4)

Want to Show True: S(k+1) = 2+7+12+...+(5(k+1)+2) = 1/2((k+1)+1)(5(k+1)+4)

Here's what I have so far.

LHS = 2+7+12+...+ (5k+2) + (5(k+1)+2) = 5(k+1)+ 2 + 1/2(k+1)(5k+4) by induction hypothesis

I am stuck here...if anyone can help I appreciate it.

This is actually a sum of n+1 terms,
so you should have 2+7=0.5(2)9, for n=1,
in other words

$\displaystyle S_{n+1}=\frac{n+1}{2}(5n+4)$

You just need to continue on from where you were...

You are trying to prove if $\displaystyle 5(k+1)+2+\frac{1}{2}(k+1)(5k+4)=\frac{1}{2}(k+1+1) (5(k+1)+4)$ ?

as you are using the "fact" that the sum of (n+1) terms should be $\displaystyle \frac{1}{2}(n+1)(5n+4)$

Here's one way to show it....

$\displaystyle \frac{1}{2}(k+1+1)(5(k+1)+4)=\frac{1}{2}(k+1)(5(k+ 1)+4)+\frac{1}{2}(5(k+1)+4)$

$\displaystyle =\frac{1}{2}(k+1)(5k+4)+\frac{1}{2}(k+1)5+\frac{1} {2}(5(k+1)+4)$

$\displaystyle =\frac{1}{2}(k+1)(5k+4)+\frac{5}{2}k+\frac{5}{2}+\ frac{5}{2}k+\frac{5}{2}+\frac{4}{2}$

$\displaystyle =\frac{1}{2}(k+1)((5k+4)+5k+5+2$ true
• Feb 23rd 2010, 06:38 PM
forumlurker
Sorry double post.
• Feb 23rd 2010, 06:41 PM
forumlurker
Yes I am trying to show that these are equal.

$\displaystyle 5(k+1)+ 2 + 1/2(k+1)(5k+4) = 1/2((k+1)+1)(5(k+1)+4)$

I understand till this point though.

Quote:

$\displaystyle \frac{1}{2}(k+1+1)(5(k+1)+4)=\frac{1}{2}(k+1)(5(k+ 1)+4)+\frac{1}{2}(5(k+1)+4)$

My teacher has advised us to basically expand $\displaystyle 5(k+1)+ 2 + 1/2(k+1)(5k+4)$ to $\displaystyle 1/2((k+1)+1)(5(k+1)+4)$ to prove they are equal.

And to stop any assumptions, this is not a homework problem, this is for me to learn. Although my instructor is helpful, he speaks quickly and is sometimes hard to understand.
• Feb 23rd 2010, 07:19 PM
Hi forumlurker,

$\displaystyle \frac{1}{2}(k+1\color{red}+1\color{black})(5(k+1)+ 4)=\frac{1}{2}(k+1)(5(k+1)+4)+\frac{1}{2}(\color{r ed}1\color{black})(5(k+1)+4)$

If you are trying to discover if both sides are equal,
it's irrelevant whether you prove RHS=LHS or LHS=RHS.

However, sometimes you run into a teacher that will stick to one way.

$\displaystyle 5(k+1)+2+\frac{1}{2}(k+1)(5k+4)$

$\displaystyle =\frac{1}{2}5(k+1)+\frac{1}{2}5(k+1)+\frac{1}{2}2+ \frac{1}{2}2+\frac{1}{2}(k+1)(5k+4)$

$\displaystyle =\frac{1}{2}(5(k+1)+4)+\color{blue}\frac{1}{2}5(k+ 1)+\frac{1}{2}(k+1)(5k+4)$

$\displaystyle \color{blue}=\frac{1}{2}(k+1)(5k+5+4)\color{black} +\frac{1}{2}(5(k+1)+4)$

$\displaystyle =\frac{1}{2}(k+1)(5(k+1)+4)+\frac{1}{2}(5(k+1)+4)$

The first term of this sum is (k+1) times the second term,
so we can write it as

$\displaystyle \frac{1}{2}(k+1+1)(5(k+1)+4)$ proven
• Feb 23rd 2010, 07:29 PM
forumlurker
Sorry I guess this is just over my head today or something because it's not connecting. What's funny is many other induction problems aren't that difficult for me.

Thanks for offering your time and help, not many people do that.
• Feb 23rd 2010, 07:39 PM
The induction process is not very complex here,
maybe it's the rearrangements of the factors you may not be used to.

$\displaystyle \frac{1}{2}(k+2)(5k+9)=\frac{1}{2}\left(5k^2+19k+1 8\right)$
$\displaystyle 5(k+1)+2+\frac{1}{2}(k+1)(5k+4)=5k+7+\frac{1}{2}\l eft(5k^2+9k+4\right)=$$\displaystyle \frac{1}{2}\left(10k+14+5k^2+9k+4\right)=\frac{1}{ 2}\left(5k^2+19k+18\right)$