Let the given statement be P(n), i.e., P (n) : a + ar + $ar^2$ + ... + $ar^{n-1}$ = $\frac{a(r^n-1)}{r-1}$ First we prove that statement is true for n = 1. P (1) : a = $\frac{a(r^1-1)}{r-1}$ = a , which is true Assume P(k) is true for some positive integer k, i.e., a + ar + $ar^2$ + ... + $ar^{k-1}$ = $\frac{a(r^k-1)}{r-1}$ ... (i) Now prove that P(k + 1) is also true. For this we have to prove that a + ar + $ar^2$ + ... + $ar^{k-1} + ar^{(k+1)-1}$ = $\frac{a(r^{k+1}-1)}{r-1}$ L.H.S. = a + ar + $ar^2$ + ... + $ar^{k-1} + ar^{(k+1)-1}$ = $\frac{a(r^k-1)}{r-1} + ar^{(k+1)-1}$ From (i) = $\frac{ar^k - a + ar^k (r-1)}{r-1}$ = $\frac{ar^k - a + ar^{k+1} - ar^k}{r-1}$ = $\frac{ar^{k+1}-a}{r-1}$ = $\frac{a(k^{k+1}-1)}{r-1}$ = R.H.S. Thus, P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction P(n) is true ∀ n ∈ N.