Prove (by induction) that the given closed form indeed gives the same sequence. Show that the sequence defined by $_=_{−1}+2$ for $≥2$, where $_1=4$, is equivalently described by the closed formula $_=2+1$. Start by writing first $6$ terms and then you’ll need to prove (by induction) that the given closed form indeed gives the same sequence.
 A: The given closed form does not give the same sequence. For instance $$b_2=b_1+2\cdot2 =4+4=8\neq \big[2n+1\big]_{n=2}=2\cdot 2+1=5.$$
To find the correct closed form we do this:
\begin{align}
b_1=&~4\\
b_2=&~b_1+4\\
b_3=&~b_2+6=b_1+4+6\\
b_4=&~b_3+8=b_2+6+8=b_1+4+6+8.
\end{align}
We now suspect that
$$b_n=b_1+\underbrace{4+6+8+\ldots+2n}_{(n+2)(n-1)}.$$
(If you want you can prove by induction that $4+6+8+\ldots+2n=(n+2)(n-1)$.) 
In the OP it says we should prove that the closed form we found is the correct one by using induction. So let us do that.  

We want to prove that the sequence $b_n$ given by $b_n=b_{n-1}+2n$, $b_1=4$ is given in closed form as $b_n=4+(n+2)(n-1)$.  
Induction basis:  We have that $b_1=4+(1+1)(1-1)=4$. So this checks out.  
Induction hypothesis: Suppose $b_n=4+(n+2)(n-1)$ for some $n>0$. 
Induction step: We have $b_{n+1}=b_n+2(n+1)$. By using the induction hypothesis we may say:
\begin{align}
b_{n+1}&=b_n+2(n+1)=\\ &=4+(n+2)(n-1)+2(n+1)=\\ &=4+n^2+n-2+2n+2=\\ &=4+n^2+3n=\\ &=4+(n+3)n.
\end{align}
Which means that  $b_{n+1}=4+(n+3)n$, as desired.
