$a_n=a_{n-1}\displaystyle \frac{n+1}{n}$ if $n > 1$

$a_n=1$ if $n=1$

I'm not too sure where to start here. This is part of a review for a class and I can't really seem to remember what we're reviewing. The first 5 values are...

$a_1 = 1,a_2=1.5,a_3=2,a_4=2.5,a_5=3$

Sums of these to each point....

$a_1=1, a_2=2.5, a_3=4.5, a_4=7, a_5=10$

It doesn't seem like it should be too tricky to figure out how to get a formula for a sum of the first N terms, since each term seems to just increase by 0.5 every team, I just haven't done this for a while and am a little rusty. Any pointers would be greatly appreciated!

  • 1
    $\begingroup$ Do you know how to sum an Arithmetic Progression? $\endgroup$ – Rijul Saini Sep 5 '12 at 15:58
  • $\begingroup$ Got it! Thank you. Answer seems to be $n/2(a_1 + a_{n})$ $\endgroup$ – Hoser Sep 5 '12 at 16:02
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    $\begingroup$ @Hoser: correct, but it is better to write $n(a_1+a_n)/2$ so it is obvious that the $(a_1+a_n)$ is in the numerator. $\endgroup$ – Ross Millikan Sep 5 '12 at 16:08
  • $\begingroup$ Yeah I see what you mean. Thanks! $\endgroup$ – Hoser Sep 5 '12 at 16:19
  • $\begingroup$ The trick is to show that $a_i$ is an arithmetic progression. $\endgroup$ – Thomas Andrews Sep 5 '12 at 16:53

Just to give a complete answer, the pattern for each term appears to be $a_n=\frac{n}{2}+\frac12$

To prove it, this correctly gives $a_1=1$ and $a_{n-1}\frac{n+1}{n}= \left(\frac{n-1}{2}+\frac{1}{2}\right)\frac{n+1}{n}=\frac{n+1}{2}=\frac{n}{2}+\frac12$ so the hypothesis is true by induction

As you say, the sum of the first $n$ terms is $\frac{n(a_1+a_n)}{2}$, since this is an arithmetic progression. This is $\frac{n\left(1+\frac{n}{2}+\frac12\right)}{2} = \frac{n\left(n+3\right)}{4}$ and for example gives the sum of the first five terms as $\frac{5 \times 8}{4} = 10$ as you found


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