Prove the following by using the principle of mathematical induction for all $n in N: rac{1}{2}+ rac{1}{4}+ rac{1}{8}+$

Solution:

Let the given statement be P(n), i.e.,

$\mathrm{P}(n): \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{n}}=1-\frac{1}{2^{n}}$

For $n=1$, we have

$P(1): \frac{1}{2}=1-\frac{1}{2^{1}}=\frac{1}{2}$, which is true.

 

Let $P(k)$ be true for some positive integer $k$, i.e.,

$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots .+\frac{1}{2^{k}}=1-\frac{1}{2^{k}}$

We shall now prove that P(k + 1) is true.

Consider

$\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots \ldots+\frac{1}{2^{k}}\right)+\frac{1}{2^{k+1}}$

$=\left(1-\frac{1}{2^{k}}\right)+\frac{1}{2^{k+1}} \quad[$ Using (i) $]$

$=1-\frac{1}{2^{k}}+\frac{1}{2.2^{k}}$

$=1-\frac{1}{2^{k}}\left(1-\frac{1}{2}\right)$

$=1-\frac{1}{2^{k}}\left(\frac{1}{2}\right)$

$=1-\frac{1}{2^{k+1}}$

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.