Prove that the formula for the sum to n term of a geometric series is given by the formula `S_n = (a(r^n-1))/(r-1)`

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The geometric series with initial term `‘a’` , common ratio `‘r’` and sum of n terms `S_n` can be written as;





Note that first term in (2) and second term in (1) cancel out. Same happens in the same way except last term in (2) and first term in (1). So finally we get;

`rS_n - S_n=ar^n-a`

`S_n(r-1) =a(r^n-1)`

`S_n =(a(r^n-1))/(r-1)`

So the answer is proved.


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