# Thread: Need help with mathematical induction.

1. ## Need help with mathematical induction.

Here's a problem from tonight's homework that I am stuck on.

$\displaystyle 3+8+13+18+...+(5n-2)=\frac{n}{2}(5n+1)$

Here's the work I did up to the point I'm stuck:

$\displaystyle s_{1}=3=\frac{1}{2}[5(1)+1]\surd$
$\displaystyle s_{k}=3+8+13+18+...+(5n-2)=\frac{k}{2}(5k+1)$
$\displaystyle s_{k+1}=3+8+13+18+...+(5n-2)+[5(k+1)-2]$
$\displaystyle =s_{k}+(5k+3)$
$\displaystyle =\frac{k}{2}(5k+1)(5k+3)$

I know that $\displaystyle s_{k+1}$ is supposed to equal $\displaystyle \frac{k+1}{2}(5k+6)$ when it's all said and done. But at this point, I cannot figure out how to reach that.

I will greatly appreciate any and all help.

2. Hi
Originally Posted by mathgeek777
$\displaystyle 3+8+13+18+...+(5n-2)=\frac{n}{2}(5n{\color{red}+}1)$
It should be easier to work with that.

3. The minus sign was put in incorrectly. But I still came out to the right conclusion on the work I gave.

Still need help folks

4. Let's evaluate $\displaystyle s_{k+1}$ : $\displaystyle s_{k+1}=s_k+(5k+3)=\frac{k}{2}(5k+1)+(5k+3)=\frac{ 5k^2+k+10k+6}{2}=\frac{5k^2+11k+6}{2}$ hence $\displaystyle \boxed{s_{k+1}=\frac{(k+1)(5k+6)}{2}}$

5. Of course it looks so obvious .

Thank you sir for the assistance. It is appreciated greatly.