To prove:
\(\frac{1}2+\frac{1}4+\frac{1}8\)+,,,,,+\(\frac{1}{2^n}\) = \((1-\frac{1}{2^n})\)
Steps to prove by mathematical induction:
Let P(n) be a statement involving the natural number n such that
(i) P(1) is true
(ii) P(k + 1) is true, whenever P(k) is true
Then P(n) is true for all n ϵ N
Therefore,
Let (Pn); \(\frac{1}2+\frac{1}4+\frac{1}8\)+,,,,,+\(\frac{1}{2^n}\) = \((1-\frac{1}{2^n})\)
Step1:
P(1) = 1 - \(\frac{1}{2^1}\) = 1 - \(\frac{1}{2}\) = \(\frac{1}{2}\)
Therefore, P(1) is true
Step 2:
Let P(k) is true Then,
P(k): \(\frac{1}2+\frac{1}4+\frac{1}8\)+,,,,,+\(\frac{1}{2^k}\) = \(1-\frac{1}{2^k}\)
Now,
\(\frac{1}2+\frac{1}4+\frac{1}8\)+,,,,,+\(\frac{1}{2^{k+1}}\) = \(1-\frac{1}{2^k}\) + \(\frac{1}{2^{k+1}}\)
= P(k + 1)
Hence, P(k + 1) is true whenever P(k) is true
Hence, by the principle of mathematical induction, we have
\(\frac{1}2+\frac{1}4+\frac{1}8\)+,,,,,+\(\frac{1}{2^n}\) = \((1-\frac{1}{2^n})\)
for all n ϵ N