$$\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + … + \dfrac{1}{2^n} = 1 – \dfrac{1}{2^n}$$

Asked by Aaryan | 1 year ago |  59

##### Solution :-

Let us check for n = 1,

P (1): $$\dfrac{ 1}{2^1} = 1 – \dfrac{1}{2^1}$$

$$\dfrac{1}{2}= \dfrac{1}{2}$$

P (n) is true for n = 1.

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

Let P (k): $$\dfrac{1}{2}+ \dfrac{1}{4}+ \dfrac{1}{8} + … + \dfrac{1}{2^k} = 1 – \dfrac{1}{2^k}$$ … (i)

So,

$$\dfrac{1}{2}+ \dfrac{1}{4}+ \dfrac{1}{8} + … + \dfrac{1}{2^k} +\dfrac{1}{2^{k+1}}$$

Now, substituting the value of P (k) we get,

= $$1 – \dfrac{1}{2^k} + \dfrac{1}{2k+1}$$ by using equation (i)

= $$1 – \dfrac{(2-1)}{2^{(k+1)}}$$

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

Answered by Aaryan | 1 year ago

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