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Using the principle of mathematical induction, prove each of the following for all n ϵ N:

\(\frac{1}2+\frac{1}4+\frac{1}8\)+ ....... +\(\frac{1}{2^n}\) = \((1-\frac{1}{2^n})\)

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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

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