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

1 Answer

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.,


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


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

\( 1-\dfrac{1}{2^k}+\dfrac{1}{2^{k+1}}\)............[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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