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Prove $a_1+\cdots+a_n=\dfrac{(a_1+a_n)n}{2}$ inductively.

Where $a_i=a_{i+1}-r$.

I tried to start proving it inductively, but any try lead to a bad conclusion, so I ended up proving it by making $a_n$ depend on $a_i$.

But I didn't know how to prove it inductively, so there is the problem.

EDIT:

I'm looking for a valid induction steps to reach the conclusion.

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  • $\begingroup$ Isn't there already a formula for finding the sum that goes from $a_1$ and $a_n$: $$a_1 +\cdots+a_n = \frac{a_n(a_n+1)}{2}$$ $$\implies \frac{a_n(a_n + 1)}{\require{cancel}\cancel{2}} = \frac{n(a_1 + a_n)}{\cancel{2}}$$ $$\implies a_n(a_n + 1) = n(a_1 + a_n)$$ $$\implies a_n \cancel{+ 1} = \frac{na_1}{a_n} \cancel{+1}$$ $$\implies a_n^2 = na_1$$ $\endgroup$ – Mr Pie Sep 12 '17 at 0:21
  • $\begingroup$ @user477343 Suppose the sum $1+2+3$. Your conclusion is that $9=3^2=3*1=3$, which is very false. It also changes the result if instead of $1+2+3$ we have $2+1+3$, and that cannot be true. $\endgroup$ – Garmekain Sep 12 '17 at 20:10
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If $n=1$, the result is trivial.

Suppose $$\sum_{i=1}^k a_i = \frac{(a_1+a_k)k}{2}$$

\begin{align}\sum_{i=1}^{k+1} a_i &= \frac{(a_1+a_k)k}{2}+a_{k+1}\\&=\frac{a_1k+a_{k+1}(k+1)-rk+a_{k+1}}{2}\\ &= \frac{a_1k+a_{k+1}(k+1)+a_1}{2}\\ &=\frac{(a_1+a_{k+1})(k+1)}{2}\end{align}

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  • $\begingroup$ bur $rk+a_{k+1}\neq a_1$ $\endgroup$ – Garmekain Mar 1 '17 at 0:03
  • $\begingroup$ You mean $-rk$ instead of $+rk$ in the third line, right? $\endgroup$ – ryanblack Mar 1 '17 at 0:39
  • $\begingroup$ @ryanblack thanks! $\endgroup$ – Siong Thye Goh Mar 1 '17 at 1:09
  • $\begingroup$ How did you get from $ka_k$ to $(k+1)a_{k+1}$? $\endgroup$ – Garmekain Mar 1 '17 at 13:08
  • $\begingroup$ $a_kk+2a_{k+1} = (a_{k+1} - r) k + a_{k+1} + a_{k+1} = a_{k+1}(k+1) - rk + a_{k+1}$ $\endgroup$ – Siong Thye Goh Mar 1 '17 at 18:03
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No induction needed ... just use a simple trick famously used by Gauss when he was 10 years old:

Take two of these series, one going from $a_1$ to $a_n$, and the other one going back from $a_n$ to $a_1$, put them under each other, and add them up by entry (that is, add the first entries of the two series, then add the second entries, etc):

$a_1 + a_2 + ... + a_{n-1} + a_n$

$a_n + a_{n-1} + ... + a_2 + a_1$

Added together gives:

$(a_1 + a_n) + (a_2 + a_{n-1}) + ... + (a_{n-1} + a_2) + (a_n+a_1)$

Now note that $a_{1+i} + a_{n-i} = a_1 + i*r + a_n - i*r = a_1 + a_n$

(put differently: each time you move an entry to the right, the first number of the pair increases by $r$, while the second of the pair decreases by $r$, so the sum stays the same)

So, each pair adds up to $a_1 + a_n$, and since you have n pairs, you get a total of $n*(a_1 + a_n)$.

Since that is the sum of two series, one series has a sum of half of that, i.e.:

$$a_1 + ... + a_n = \frac{n*(a_1 + a_n)}{2}$$

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$a_1+a_2={2{a_1+a_2\over2}}$. You have $a_2=a_1+r$ and recursively $a_n=a_1+(n-1)r$, $a_1+...+a_n=a_1+r+a_1+2r+...+a_1+(n-1)r=na_1+r{n(n-1)\over 2}$ =${{na_1+na_1+rn(n-1)}\over 2}$ =$n{{a_1+a_1r(n-1)}\over 2}=n{{a_1+a_n}\over 2}.$

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  • $\begingroup$ This is what I meant saying that I defined $a_n$ as function of $a_1$. $\endgroup$ – Garmekain Mar 1 '17 at 0:04
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Statement true for $n=2$. Assume

$$ a_1 + \dots + a_n = \frac{(a_1+a_n)n}{2}.$$

Then, using the inductive hypothesis and adding and subtracting terms, we get

$$a_1 + \dots+ a_n + a_{n+1} = \frac{(a_1+a_n)n}{2}+ a_{n+1}$$

$$= \frac{(a_1+ a_{n+1})(n+1) }{2} + \frac{a_nn-a_1}{2}+ a_{n+1} - \frac{a_{n+1}(n+1)}{2}. $$

Finally,

$$ \frac{a_nn-a_1}{2}+ a_{n+1} - \frac{a_{n+1}(n+1)}{2} = \frac{na_n-a_1+a_{n+1}-na_{n+1}}{2}$$

$$ = \frac{a_{n+1}-a_1-nr}{2}=0$$

since $a_{n+1} = a_1 + nr$.

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  • $\begingroup$ How did you get $\frac{(a_1+ a_{n+1})(n+1) }{2} + \frac{a_nn-a_1}{2}+ a_{n+1} - \frac{a_{n+1}(n+1)}{2} $? $\endgroup$ – Garmekain Mar 1 '17 at 0:33
  • $\begingroup$ I simply aded and subtracted zero. Or, if you prefer, the following $$\frac{a_n n -a_1}{2}+a_{n+1} -\frac{a_{n+1}(n+1)}{2} =\left( \frac{(a_1+a_n)n}{2}+a_{n+1}\right) - \left( \frac{(a_1+a_{n+1})(n+1)}{2}\right)$$ is just an algebraic identity: no induction or anything else needed. $\endgroup$ – Stefano Mar 1 '17 at 9:30

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