$a_n=a_{n-1}\displaystyle \frac{n+1}{n}$ 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!
Best Answer
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