Prove by induction: $\sum\limits_{i=1}^{n}(4i+1) = 2n^2 + 3n$

Question

Prove by induction: $$\sum\limits_{i=1}^{n}(4i+1) = 2n^2 + 3n$$

It's just the numbers that confuse me; I know how to do a simple induction proof that first for $p(k)$ and then for $k+1$ etc but I don't know how to do it with the form?

Answer 1

If you want to make it by induction. Look what you have and what you want, i.e.

You have $$\color{blue}{\sum\limits_{i=1}^{n}(4i+1) = 2n^2 + 3n}$$ and you want $$\sum\limits_{i=1}^{n+1}(4i+1) = 2(n+1)^2 + 3(n+1).$$

So write $$\sum\limits_{i=1}^{n+1}(4i+1)=(4(n+1)+1)+\color{blue}{\sum\limits_{i=1}^{n}(4i+1)}$$ use what you have and try to recover $2(n+1)^2 + 3(n+1)$.

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Here is also a proof without induction:(or maybe very little induction).

Note that $$\sum_{i=1}^n (4i+1)=4\left(\sum_{i=1}^ni \right) + \sum_{i=1}^n1=...$$

and $$2\sum_{i=1}^n i =(1+2+\ldots+n) + (1+2+\ldots+n)\\=(1+n)+(2+(n-1))+\ldots+(n+1)=n(n+1)$$

So that $\sum_{i=1}^ni=\frac{n(n+1)}{2}$, noting that $\sum_{i=1}^n1=n$, you should be able to conclude.

Answer 2

You need to show that $\sum_{i=1}^{n+1}4i+1=2(n+1)^2+3(n+1)=2n^2+7n+5$, but

\begin{align} \sum_{i=1}^{n+1}4i+1 & =\sum_{i=1}^n4i+1 +4(n+1)+1\\ & =2n^2+3n+4(n+1)+1\\ & =2n^2+7n+5 \end{align} where the second equality comes from the induction hypothesis.

Answer 3

If one doesn't like to prove the same thing by induction over and over again:

prove $$\sum\limits_{i=1}^n i=\frac{n(n+1)}2$$ by induction so you have $$\sum\limits_{i=1}^n (4i+1)=4\sum\limits_{i=1}^n i+\sum\limits_{i=1}^n 1 = 4\cdot \frac{n(n+1)}{2} + n = 2n(n+1)+n=2n^2+3n.$$

To generalize you can also calculate $$\sum_{i=1}^n (ai+b), a,b\in\mathbb R$$ as $$\sum_{i=1}^n (ai+b)=\frac 12n(an+a+2b).$$ For $a=4,b=1$ we get $$\sum\limits_{i=1}^n (4i+1)=\frac 12n(4n+4+2)=2n^2+3n.$$