Prove the following by using the principle of mathematical induction for all `n in N`:`a+a r+a r^2+dotdotdot+a r^(n-1)=(a(r^n-1))/(r-1)`

Prove the following by using the principle of mathematical induction f

1.	Prove the following by using the principle of mathematical induction for all `n in N`:`a+a r+a r^2+dotdotdot+a r^(n-1)=(a(r^n-1))/(r-1)`
Answer» Let the given statement be P(n) . Then , `P(n): a + ar + ar^(2) + …+ar^(n-1)= (a(r^(n)-1))/((r-1))`. When n = 1, we have LHS = a and RHS = `(a(r^(1)-1))/((r-1))`. When n = 1, we have LHS = a and RHS = `(a(r^(1)-1))/((r-1)) = a`. `:. `LHS = RHS. Thus, P(1) is true. LetP(k) be true. Then, `P(k) : a+ar+ar^(2)+...+ar^(k-1)=(a(r^(k)-1))/((r-1))`. ...(i) Now, `(a+ar+ar^(2)+...ar^(k-1)) + ar^(k) = (a(r^(k)-1))/((r-1)) + ar^(k)` [using (i)] `=(a(r^(k+1)-1))/((r-1))`. `:. P(k+1): a+ar+ar^(2)+...+ar^(k-1)+ar^(k) = (a(r^(k+1)-1))/((r-1))`. This shows that P(k+1) is true, whenever P(k) is true. `:. " "` P(1) is true and P(k+1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, we have ` a+ar+ar^(2) +...+ ar^(n-1) = (a(r^(n)-1))/((r-1)) " for " r gt 1 and " all "n in N`.