\(\displaystyle \frac{1}{2} + \frac{1}{4}+ \frac{1}{8} + ......... + \frac{1}{2^n} = 1 - \frac{1}{2^n}\) 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

(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

Answer: D

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

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

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