Question

Prove the following by using the principle of mathematical induction for all $n\in N$
$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{2^n}=1-\frac{1}{2^n}$

Answer 1

Step $\left(1\right):$ Assume given statement
Let the given statement be $P\left(n\right)$, i.e.,
$P\left(n\right):\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{2^n}=1-\frac{1}{2^n}$

Step $\left(2\right):$ Checking statement $P\left(n\right)$ for $n=1$
Put $n=1$ in $P\left(n\right)$, we get
$P\left(1\right):\frac{1}{2}=1-\frac{1}{2^1}$
$\Rightarrow \frac{1}{2}=\frac{1}{2}$
Thus $P\left(n\right)$ is true for $n=1$

Step $\left(3\right):$ $P\left(n\right)$ for $n=K$
Put $n=K$ in $P\left(n\right)$ and assume this is true for some natural number $K$ i.e.,
$P\left(K\right):\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{2^K}=1-\frac{1}{2^K}\cdots \left(1\right)$

Step $\left(4\right):$ Checking statement $P\left(n\right)$ for $n=K+1$
Now we shall prove that $P(K+1)$ is true whenever $P\left(K\right)$ is true.
Now, we have
$P\left(K\right):\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{2^K}+\frac{1}{2^{K+1}}=1-\frac{1}{2^K}+\frac{1}{2^{K+1}}$ (Using $\left(1\right)$)
$=1+\frac{1}{2^{K+1}}-\frac{1}{2^K}$
$=1+\frac{1-2}{2^{K+1}}$
$=1-\frac{1}{2^{K+1}}$
Thus, $P(K+1)$ is true whenever $P\left(K\right)$ is true.
Final Answer :
Hence, from the principle of mathematical induction, the statement $P\left(n\right)$ is true for all $n\in N$.