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I don't remember how to solve these, but I hope someone can refresh my memory. I have the following equation, how can I solve for t?

$$50000 = \sum_{i=0}^{12t} 1.004^{12(t - i/12)}$$

EDIT: if it helps at all, expands out to $$50000 = 1.04^{12t} + 1.04^{12(t-1/12)} + 1.04^{12(t-2/12)} + .... + 1$$

holtc
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    Can you find a geometric series in there? – GEdgar Oct 02 '15 at 15:51
  • Normally, I input t in order to find the 50000, but now I am trying to do it in reverse, given the 50000, find t – holtc Oct 02 '15 at 15:55
  • Numerically, $t=110.623...$ – Zach466920 Oct 02 '15 at 16:26
  • Second hint to support @GEdgar's one: Show that $$1.04^{12t} + 1.04^{12(t-1/12)} + 1.04^{12(t-2/12)} + \cdots + 1$$ is actually $$1.04^{12t} + 1.04^{12t-1} + 1.04^{12t-2} + \cdots + 1$$ that is, with $r=1.04$, $$\sum_{i=0}^{12t}r^i=\frac{r^{12t+1}-1}{r-1},$$ hence... – Did Oct 02 '15 at 16:56
  • so $t$ won't be an integer as it seems, how should your sum then actually work? – user190080 Oct 02 '15 at 17:17
  • @user190080 In case your comment is directed at me: 1. please use @, 2. I have no idea what you are talking about (of course $12t$ should be an integer). – Did Oct 03 '15 at 16:37
  • @Did no, the comment was meant to address the OP...to clarify: if you solve it the way you did and as in the accepted answer, you get a non-integer $t$, but this of course means that the sum is sort of not well defined if you just plug in $t$. so it won't work out as the OP thought (like the last term of summation is $1$) and in the long run, the equality simply doesn't hold for any integer $t$. – user190080 Oct 03 '15 at 17:05

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$$\begin{align} 50000 &= \sum_{i=0}^{12t} 1.004^{12(t - i/12)} \\ &=\sum_{i=0}^{12t} 1.004^{12t - i} \\ &=\sum_{k=0}^{12t} 1.004^k\qquad\text{where $k=12t-i$} \\ &=\frac{1.004^{12t+1}-1}{1.004-1} \\ 1.004^{12t+1}&=201 \\ t&=\frac 1{12}\left(\frac{\ln 201}{\ln1.004}-1\right)=110.62\qquad\blacksquare \end{align}$$

Hypergeometricx
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