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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!

Ana
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    You should write this task here, no paste picture. It is hard to read. For some basic information about writing math at this site see e.g. here, here, here and here. – Tacet Dec 01 '14 at 21:59
  • They are really quite different from each other. Your 20a is an arithmetic progression, $n$ terms, common difference $2$.You can also solve it by noting that the sum is $2(1+\cdots+n)+4n$. Now if you know the sum of the first $n$ natural numbers, you are close to the end. – André Nicolas Dec 01 '14 at 22:00
  • For the part with $i$, see this –  Dec 01 '14 at 23:04

2 Answers2

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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.

egreg
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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.

ml0105
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