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Prove the riemann sum to the nth term with i=1 of ar^i-1 is equivalent to a(r^n -1)/(r-1).
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It remains true as a sum of a finite series with r not equal to 1 according to the formula.
so for i=1 you get ar^(1-1) = ar^0 = a and on the other part for n=1 a((r^1 -1)/(r-1)) = a((r-1)/(r-1)) = a*1 = a sorry here a(r^(n-1))/(r-1) will be right or a((r^n -1)/(r-1)) or this is right ? suppose is true for i=k and n=k than prove for k=k+1 for i and n so than will get ar^(k+1) = ar^k *r and on the other part a((r^(k+1) -1)/(r-1)) = a(r^k *r -1)/(r-1)) =
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