Question

$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.........+\frac{1}{2^n}=1-\frac{1}{2^n}$ is true for

• A. Only natural number n $\ge$ 3
• B. Only natural number n $\ge$ 5
• C. Only natural number n < 10
• D. All natural number n

Answer 1

The correct option is D All natural number n

Let P(n)$:\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+..........+\frac{1}{2^n}=1-\frac{1}{2^n}$
Putting $n=1,LHS=\frac{1}{2},RHS=1-\frac{1}{2}=\frac{1}{2}$ i.e., LHS $=$ RHS $=\frac{1}{2}$
$\therefore$ P(n) is true for n$=$ 1.
Suppose P(n) is true for n$=$ k
$\therefore \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.........+\frac{1}{2^k}=1-\frac{1}{2^k}$
last term $=\frac{1}{2^k}$; Replacing k by k+1, last term $=\frac{1}{2^{k+1}}$
Adding $\frac{1}{2^{k+1}}$ to both sides,
$LHS=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+........+\frac{1}{2^k}+\frac{1}{2^{k+1}}$
$RHS=1-\frac{1}{2^k}+\frac{1}{2^{k+1}}=1-\frac{1}{2^k}(1-\frac{1}{2})=1-\frac{1}{2^k}\cdot \frac{1}{2}=1-\frac{1}{2^{k+1}}$
This shows P(n) is true for n $=$ k+1
Thus P(k+1) is true whenever P(k) is true
Hence, P(n) is true for all n $\in$ N