How to find closed form by induction

How can I find the closed form of

a) 1+3+5+...+(2n+1)
b) 1^2 + 2^2 + ... + n^2

using induction?

I'm new to this site, and I've thought about using the series 1 + 2 + 3 +...+ n = n(n+1)/2 to help me out But isn't that technically using prior knowledge and hence invalid? Am I on the right track? Thanks.

Answer

$a):$ Clearly it is an Arithmetic Series

So, the sum up to $n$th term is $=\frac n2\{2\cdot1+(n-1)2\}=n^2$

Let $P(n):\sum_{1\le r\le n}(2r-1)=n^2$

$P(1):\sum_{1\le r\le 1}(2r-1)=1$ which is $=1^2$ so $P(n)$ is true for $n=1$

Let $P(n)$ is true for $n=m$

So, $\sum_{1\le r\le m}(2r-1)=m^2$

$P(m+1): \sum_{1\le r\le m+1}(2r-1)=\sum_{1\le r\le m}(2r-1)+2m+1=m^2+2m+1=(m+1)^2$

So, $P(m+1)$ will be true if $P(m)$ is true.

But we have already shown $P(n)$ is true for $n=1$

So, by induction we can prove that $P(n)$ is true for all positive integer $n$

$b):$

HINT:

We know, $(r+1)^3-r^3=3r^2+3r+1$

Putting $r=1,2,\cdots,n-1,n$ and adding we get $(n+1)^3-1=3\sum_{1\le r\le n}r^2+3\sum_{1\le r\le n}r+\sum_{1\le r\le n}1$

Now, you know $\sum_{1\le r\le n}r=\frac{n(n+1)}2$ and $\sum_{1\le r\le n}1=n$

On simplification $\sum_{1\le r\le n}r^2=\frac{n(n+1)(2n+1)}6$

Can you use similar induction approach here?