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Here is a geometric proof of why $\sum ar^{n}=\dfrac{a}{1-r}$ (|r|<1)-Geometric Proof

which I find is quite intuitive.

Can someone give me a hint to prove $\sum_{k=0}^{n-1} ar^{k}=\dfrac{a(1-r^{n})}{1-r}$ (k is finite,|r|<1) using similar geometric interpretation.

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up vote 2 down vote accepted

The geometrical argument linked to works for $0\lt r\lt 1$. It can be used to find the partial sum $\sum_{k=0}^{n-1} ar^k$. Since the argument will use the infinite sum, again the geometry works only for $0\lt r\lt 1$.

The (infinite) sum, from the beginning to the end, is $\frac{a}{1-r}$. And the tail, where we sum from $n$ to infinity, is $\frac{a}{1-r}$, scaled by the factor $r^n$. The difference between the whole sum and the tail is what we were after. It is $$\frac{a}{1-r}-r^n\frac{a}{1-r},$$ exactly the desired result.

Remark: An amusing reversal! We ordinarily use the limit of partial sums to find the infinite sum. Here we used the geometry of the infinite sum to calculate the partial sum.

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Brilliant.Thank you! –  shaswata Apr 24 '13 at 16:20
    
@shaswata: You are welcome. Scaling arguments are very useful in mathematics, and even more so in Physics. –  André Nicolas Apr 24 '13 at 16:23
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