Use mathematical induction to prove $\sum_{i=1}^{n}(2i+4)=n^2+5n$

Question

Prove: $$ \sum_{i=1}^{n}(2i+4)=n^2+5n \textrm{ for each positive integer } n $$

So I'm not exactly sure how to do this problem for my math class. Can any mathematicians out there help me? Explaining how to do it is good. I should be able to do the rest once I have the first one figured out. Thank you!

Answer 1

Hint- break up the series:

$$\sum_{i=1}^{n} (2i + 4) = 2\sum_{i=1}^{n} i + \sum_{i=1}^{n} 4$$

Have you already shown that $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$? Then you can note that $2n(n+1)/2 = n(n+1)$. And then you have $\sum_{i=1}^{n} 4 = 4n$. And so you get $n^{2} + n + 4n = n^{2} + 5n$.

If you haven't already shown that $\sum_{i=1}^{n} i = n(n+1)/2$, do that by induction. Then the rest of your proof comes from breaking terms up and factoring.

Answer 2

You can do it by standard induction; consider the sequence of statements $$ P(n){:}\qquad \sum_{i=1}^n(2i+4)=n^2+5n $$ Then $P(1)$ reads $2\cdot 1+4=1^2+5\cdot1$ which is true.

Suppose $P(n)$ holds; then you want to prove that $$ \sum_{i=1}^{n+1}(2i+4)=(n+1)^2+5(n+1)=n^2+7n+6 $$ By assumption, $\sum_{i=1}^n(2i+4)=n^2+5n$, so $$ \sum_{i=1}^{n+1}(2i+4)= \biggl(\sum_{i=1}^{n}(2i+4)\biggr)+(2(n+1)+4)= n^2+5n+2n+2+4=n^2+7n+6 $$ and we are done.