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Suppose $$X = 1.05^{35}v+1.05^{34}v^{2} + \cdots + 1.05v^{35}$$ where $v = 1/1.05$. Then we have $$X = 1.05^{35}v(1+ 0.952v+ \cdots + 0.952^{34}v^{34})$$

So the sum of this would be $$ X = 1.05^{35}v \left[\frac{1-\left(\frac{1+952v}{1.05} \right)^{35}}{1-\frac{1+.952v}{1.05}} \right]$$

Is that right?

  • 6
    Except that $1/1.05\ne .952$. (this will be important when taking a big power)2011-12-04
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    So, to clarify, the sum in question is $X=\sum_{i=1}^{35} 1.05^{36-i}\left(\frac{1}{1.05}\right)^{i}$ ?2011-12-04
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    Additionally, what is the context of this problem? It looks closest to finding the PV of a growing annunity, but its not exactly the same.2011-12-04

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