Prove the following by using the principle of mathematical induction for all $$n\in N$$

$$\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}...+\dfrac{1}{2^n}=1-\dfrac{1}{2^n}$$

Asked by Abhisek | 1 year ago |  89

##### Solution :-

Let us denote the given equality by P(n), i.e.,

$$P(n): \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}...+\dfrac{1}{2^n}=1-\dfrac{1}{2^n}$$

For n=1,

L.H.S.= $$\dfrac{1}{2}$$

R.H.S.$$1- \dfrac{1}{2^1}=\dfrac{1}{2}$$

Therefore, P(n) is true for n=1.

Let us assume that P(k) is true for some positive integer k , i.e.,

$$\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}...+\dfrac{1}{2^k}=1-\dfrac{1}{2^k}$$.........(i)

Now, we have to prove that P(k+1) is also true.

Consider

$$( \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}...+\dfrac{1}{2^k})+\dfrac{1}{2^{k+1}}$$

$$1-\dfrac{1}{2^k}+\dfrac{1}{2^{k+1}}$$............[Using (i)]

$$1-\dfrac{1}{2^k}( 1-\dfrac{1}{2})$$

$$1-\dfrac{1}{2^k}( \dfrac{1}{2})$$

Therefore, P(k+1) holds whenever P(k) holds.

Hence, the given equality is true for all natural numbers i.e., N by the principle of mathematical induction.

Answered by Pragya Singh | 1 year ago

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