How Do Telescoping Sums Relate to Squares in Mathematical Proofs?

nameVoid
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the proof in my text starts with what's called a telescoping sum (1+i^3)-i^3 what is the relevence of this to i^2
 
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a) Prove the formula for the partial geometric series:

\sum_{k=0}^{n}x^{k}=\frac{1-x^{n+1}}{1-x}


Hint: Multiply both sides by 1-x.


b) Substitute x = e^t in the formula for the partial geometric series.


c) Perform a series expansion of both sides to second order in t.
 
nameVoid said:
the proof in my text starts with what's called a telescoping sum (1+i^3)-i^3 what is the relevence of this to i^2

Shouldn't that be (1+i)^3 - i^3?

If you expand (i+1)^3, you see that i^3 cancels and then you get a combinaton of the summaton of i^2 and i and of 1.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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