Prove a formula in terms of n: $1+5+9+...+(4n+1)$
I HAVE to use induction, but I am new to induction, so I am a bit confused...
I believe I have to use the base case first: so $n=1$ is $4(1)+1=5$, but i get the second term in the sequence instead of the first term? is this ok? Next I tried to substitute k+1 in the formula and got: $4(k+1)+1=4k+5$. Is my approach feasible or what do I do next? 
Thanks!
Edit: I realise the formula would be $n+4$... but how would I go about proving it and explaining it using induction? 
 A: You may know the formula $1+2+\cdots+n=\frac{n(n+1)}{2}$. By this formula we get
$$
\begin{aligned}
1+(4+1)+(4\cdot 2+1)+\cdots + (4n+1)
&=4(1+2+\cdots +n )+(n+1)
\\&=2n(n+1)+(n+1)
\\&=(2n+1)(n+1).
\end{aligned}
$$
So desired formula is $(2n+1)(n+1)$. If you want to prove by induction, First you check $1+5=(2+1)(1+1)$ and you check $1+5+9+\cdots+(4n+1)=(n+1)(2n+1)$ implies $1+5+9+\cdots+(4n+1)+(4n+5)=(n+2)(2n+3)$.
A: Draw 5 dots on a paper and connect them with drawing segments between them.  You'll notice that $4+3+2+1$ how many lines you draw (try it!).  But luckily we have another way of counting this using the binomial coefficient, and that's $5 \choose 2$, So in general, ${\sum_1^n j} = {n+1 \choose 2} = n(n+1)/2$.  So if you can prove the formula by induction then you'll be okay.  Try to look at some other examples of it being used. I was hoping this answer would be helpful because it shed some light on the formula mentioned in the previous answer..
A: Induction is targeted at verifying a proposition.
Let us establish the Proposition 
Using the summation formula of Arithmetic progression 
$$1+5+9+\cdots+(4n+1)=\frac{n+1}2\{1+(4n+1)\}=(n+1)(2n+1)$$
Let $f(n):1+5+9+\cdots+(4n+1)=(n+1)(2n+1)$ 
Clearly, $f(n)$ is true for $n=1$
Let $f(n)$ is true for $n=m\implies 1+5+9+\cdots+(4m+1)=(m+1)(2m+1)=2m^2+3m+1$
Putting $n=m+1,$
$\implies 1+5+9+\cdots+(4m+1)+\{4(m+1)+1\}=2m^2+3m+1+(4m+5)$ using $f(m)$
$\implies 1+5+9+\cdots+(4m+1)+\{4(m+1)+1\}=2m^2+7m+6=2m^2+4m+3m+6=2m(m+2)+3(m+2)=(m+2)(2m+3)=\{(m+1)+1\}\{2(m+1)+1\}$
So, $f(n)$ holds true for $n=m+1$ if  $f(n)$ holds true for $n=m$
