Using the method of mathematical induction, show that for all n∈N,

1.	Using the method of mathematical induction, show that for all n∈N, a + ar + ar2 + ... +arn – 1 = \(\frac{a(1-r^n)}{(1-r)}\) , r ≠ 1.
Answer» Let T(n) be the statement: a + ar + ar² + ... + ar^n–1 = \(\frac{a(1-r^n)}{(1-r)}\) , r ≠ 1 Basic Step: For n = 1, ⇒LHS = a, RHS = \(\frac{a(1-r^1)}{(1-r)}\)= a. ⇒LHS = RHS ⇒ T(1) is true. Induction Step: Let the statement hold true for n = k, i.e., let T (k) be true, i.e., a + ar + ar² + ... + ar^k–1 = \(\frac{a(1-r^k)}{(1-r)}\) Then to show T(k + 1) holds, add the (k + 1)^th term = ar^{(k + 1)–1} = ar^k to both the sides of T(k), i.e., a + ar + ar² + ... + ar^{k – 1} + ar^k =\(\frac{a(1-r^k)}{(1-r)}\) + ar^k \(\frac{a-ar^k+ar^k(1-r)}{(1-r)} = \frac{a-ar^k+ar^k-ar^{k+1}}{(1-r)}\)= \(\frac{a-ar^{k+1}}{(1-r)}\) = \(\frac{a(1-r^{k+1})}{(1-r)}\) Thus, T(k + 1) is true, whenever T(k) holds true.

Using the method of mathematical induction, show that for all n∈N, a + ar + ar2 + ... +arn – 1 = \(\frac{a(1-r^n)}{(1-r)}\) , r ≠ 1.

Answer»

Let T(n) be the statement:

a + ar + ar² + ... + ar^n–1 = \(\frac{a(1-r^n)}{(1-r)}\) , r ≠ 1

Basic Step:

For n = 1, ⇒LHS = a, RHS = \(\frac{a(1-r^1)}{(1-r)}\)= a.

⇒LHS = RHS ⇒ T(1) is true.

Induction Step:

Let the statement hold true for n = k, i.e., let T (k) be true, i.e., a + ar + ar² + ... + ar^k–1 = \(\frac{a(1-r^k)}{(1-r)}\)

Then to show T(k + 1) holds, add the (k + 1)^th term = ar^{(k + 1)–1} = ar^k to both the sides of T(k), i.e.,

a + ar + ar² + ... + ar^{k – 1} + ar^k =\(\frac{a(1-r^k)}{(1-r)}\) + ar^k

\(\frac{a-ar^k+ar^k(1-r)}{(1-r)} = \frac{a-ar^k+ar^k-ar^{k+1}}{(1-r)}\)= \(\frac{a-ar^{k+1}}{(1-r)}\) = \(\frac{a(1-r^{k+1})}{(1-r)}\)

Thus, T(k + 1) is true, whenever T(k) holds true.