Question 6

If $$\left(\frac{1}{2^1}\right) + \left(\frac{1}{2^2}\right) + \left(\frac{1}{2^3}\right) ....... \left(\frac{1}{2^{10}}\right) = \frac{1}{k}$$, then what is the value of $$k$$?

Solution

$$\left(\frac{1}{2^1}\right)+\left(\frac{1}{2^2}\right)+\left(\frac{1}{2^3}\right).......\left(\frac{1}{2^{10}}\right)=\frac{1}{k}$$

or, $$\left(\frac{1}{2}\right)\ \frac{\left(1-\frac{1}{2^{10}}\right)}{\left(1-\frac{1}{2}\right)}=\frac{1}{k}\ .$$

or, $$\frac{1024-1}{1024}=\frac{1}{k}\ .$$

or, $$k=\frac{1024}{1023}.$$

B is correct choice.

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