Question

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

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

Answer 1

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.