Prove (by induction) that the given closed form indeed gives the same sequence.

Question

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.

Answer 1

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.