By using mathematical induction prove that $1+ a + a^2+....a^n =\dfrac{1-a^{n+1}}{1-a}$ , where n>=0.

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written 9.0 years ago by

teamques10 ★ 70k

modified 3.8 years ago by

pedsangini276 • 4.8k

Mumbai University > Computer Engineering > Sem 3 > Discrete Structures

Marks: 6 Marks

Year: May 2016

discrete mathematics

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written 9.0 years ago by

teamques10 ★ 70k

• modified 9.0 years ago

Step1:

Let n = 0. Then, the left hand side will be

LHS = 1... 1 = 1, and

RHS = $\dfrac{1-a^{n+1}}{1-a} = [1 - a] / [1 - a] = 1$

so the formula holds true for n = 0

Step2:

Assume the formula holds true for n = …

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