# How to find value of i when ∑ from k=1 to i is defined by a recursive formula and equals 982?

Thanks for the pointers! Here's updated and edited question I'm trying to find the number of days it takes to reach 982 miles when you start traveling at 18 miles/day and decrease your speed by 2% each day. Here's the recursive formula I've put together: $$a_n={18,n=1}$$ and $$a_n=\{a_n-_1-.02*a_n-_1 , n>1\}$$

Therefore the series used to determine distance traveled is: $$\sum_{n=1} ((a_n-_1)) -(.02*a_n-_1))=982$$

How do I use this with to find the day at which the 982 mile-mark is reached? Thanks!

• Its not clear what your asking here. Can you clean up the notation a bit or add some more explanation of what the problem is? – Spencer Apr 28 '15 at 22:13
• Here is a through guide for formatting mathematics on this site. Please use it to make your expressions readable. For the question itself: your sequence is of a special kind that has a name: a geometric sequence. Do you know any formulas concerning such sequences that you think might be relevant? – Arthur Apr 28 '15 at 22:17
• yes, a/(1-r) and (a_n+_1)/(a_n) – jackson Apr 28 '15 at 23:23

Let $r=2\%=\frac{2}{100}$
First day: $$a_{0}=18$$Second day: $$a_{1}=18-18r=a_{0}-a_{0}r=a_{0}(1-r)$$Third day: $$a_{2}=a_{1}-a_{1}r=a_{1}(1-r)=a_{0}(1-r)^{2}$$ Therefore on n-th day: $$a_{n}=a_{0}(1-r)^{n}$$ So we have: $$\sum_{n=0}^{N} a_{0}(1-r)^{n}=\sum_{n=0}^{N}a_{0}k^{n}$$ This is called geometric series. As $N$ goes to infinity, the absolute value of $k$ must be less than one for the series to converge i.e. $|k|<1$;
$$k=1-\frac{2}{100}=\frac{98}{100}<1$$ so condition is satisfied. It's convenient for this problem to check what happens when $N\rightarrow\infty$. $$\sum_{n=0}^{\infty}a_{0}k^{n}=\frac{a_{0}}{1-k}=\frac{18}{1-\frac{98}{100}}=900$$