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## QuestionMathsClass 11

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

(A) Only natural number n $$\geq$$ 3
(B) Only natural number n $$\geq$$ 5
(C) Only natural number n < 10
(D) All natural number n

Let P(n)$$: \displaystyle \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + .......... + \frac{1}{2^n} = 1 - \frac{1}{2^n}$$
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
$$\therefore \displaystyle \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ......... + \frac{1}{2^k} = 1 - \frac{1}{2^k}$$
last term $$= \displaystyle \frac{1}{2^k}$$; Replacing k by k+1, last term $$=\displaystyle \frac{1}{2^{k+1}}$$
Adding $$\displaystyle \frac{1}{2^{k+1}}$$ to both sides,
$$LHS = \displaystyle \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ........ + \frac{1}{2^k}+ \frac{1}{2^{k+1}}$$
$$RHS = 1 - \displaystyle \frac{1}{2^k} + \frac{1}{2^{k+1}} \left ( 1 - \frac{1}{2} \right ) = 1 - \frac{1}{2^k} \cdot \frac{1}{2} = 1 - \frac{1}{2^{k+1}}$$
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