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Prove by induction that $$a_{1}+a_{2}+...+a_{n}=\dfrac{(a_{1}+a_{n})n}{2}, \forall n\in \mathbb N$$

HINT: Supose that: $a_{i+1}-a_{i}= r, \forall i\in \mathbb N$

Let $P(n)$ be the proposition we want to prov, ie: $a_{1}+a_{2}+...+a_{n}=\dfrac{(a_{1}+a_{n})n}{2}, \forall n\in \mathbb N$

For $P(1)$ we have:$a_{1}=\dfrac{(a_{1}+a_{1})1}{2} \implies a_{1}=\dfrac{2a_{1}}{2} \implies a_{1}=a_{1}$, so $P(1)$ is true.

For $P(2)$ we have:$a_{1}+a_{2}=\dfrac{(a_{1}+a_{2})2}{2} \implies a_{1}+a_{2}=a_{1}+a_{2} $, so $P(2)$ is true.

Inductive Hypothesis: Let $n=k$ and we assume that $$P(k):=a_{1}+a_{2}+...+a_{K}=\dfrac{(a_{1}+a_{k})k}{2}$$ is true.

Inductive Step:

$$a_{1}+a_{2}+...+a_{k}+a_{k+1}=\dfrac{(a_{1}+a_{k})k}{2}+a_{k+1}$$ $$=\dfrac{1}{2}[(a_{1}+a_{k})k+2a_{k+1}]$$ Using the HINT, $a_{i+1}-a_{i}= r \implies $$a_{k}=a_{k+1}-r$ $$=\dfrac{1}{2}[(a_{1}+a_{k})k+2a_{k+1}]=\dfrac{1}{2}[(a_{1}+a_{k+1}-r)k+2a_{k+1}]$$ $$=\dfrac{1}{2}(ka_{1}+ka_{k+1}-kr+2a_{k+1})$$

How can i finish this?

Victor F Salazar G
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    Is the hint an actual hyphothesis? – dani_s Jan 10 '14 at 00:28
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    It seems to be you mean the sequence $;a_1,a_2,...;$ is an arithmetic sequence , otherwise the claim is false, of course. For example, $;1+3+4\neq\frac{(1+4)3}2;$ ... This must be added in to the given data. – DonAntonio Jan 10 '14 at 00:28
  • Also I noticed that in your solutions you always check the value of $P(2)$; in most cases it is not necessary – dani_s Jan 10 '14 at 00:32
  • The arithmetic series condition must be given. – MathArt Jul 20 '17 at 06:18

1 Answers1

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I will assume that you are given an arithmetic series. You can finish it by realizing that $a_{k + 1} = a_1 + kr$ (although it is obvious, proving this rigorously takes a quick use of induction again). Replace one of the $a_{k + 1}$ using this expression. Continuing from where you left, you will get the following $$ = \frac{1}{2}(ka_1 + ka_{k + 1} - kr + 2a_{k + 1}) \\ = \frac{1}{2}(ka_1 + ka_{k + 1} - kr + a_{k + 1} + (a_1 + kr)) \\ = \frac{1}{2}(a_1(k + 1) + a_{k + 1}(k + 1)) \\ = \frac{(a_1 + a_{k + 1})(k + 1)}{2}. $$

Pratyush Sarkar
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