Using the principle of mathematical induction, prove each of the following

Solution:

To Prove:

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

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 $n \in N$

Therefore,

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

Step 1:

$P(1)=1-\frac{1}{2^{1}}=1-\frac{1}{2}=\frac{1}{2}$

Therefore, P(1) is true

Step 2:

Let P(k) is true Then,

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

Now,

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

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

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

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

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

$=\mathrm{P}(\mathrm{k}+1)$

Hence, P(k + 1) is true whenever P(k) is true

Hence, by the principle of mathematical induction, we have

$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots .+\frac{1}{2^{n}}=\left(1-\frac{1}{2^{n}}\right)$ for all n ϵ N