Question 64

The value of $$1 + \left(1 + \frac{1}{3}\right)\frac{1}{4} + \left(1 + \frac{1}{3} + \frac{1}{9}\right)\frac{1}{16} + \left(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}\right)\frac{1}{64} + -------$$ is


The given sequence can be written as:

$$1\left(1+\ \frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...\right)+\frac{1}{3}\left(\frac{1}{4}+\frac{1}{16}+...\right)+\frac{1}{9}\left(\frac{1}{16}+\frac{1}{64}+...\right)+..$$

We know that the sum of an infinite G.P. is $$\dfrac{a}{1-r}$$, where a is the first term and r is the common ratio.

=> The first term = $$\frac{1}{1-\frac{1}{4}}=\dfrac{4}{3}$$

=> The second term = $$\frac{1}{3}\left(\frac{\left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)}\right)=\dfrac{1}{9}$$

=> The third term = $$\frac{1}{9}\left(\frac{\left(\frac{1}{16}\right)}{1-\left(\frac{1}{4}\right)}\right)=\dfrac{1}{108}$$

Observing these three terms, we see that they are in G.P. with a common ratio of $$\dfrac{1}{12}$$

=> Sum of this infinite G.P. = $$\dfrac{\left(\dfrac{4}{3}\right)}{1-\left(\dfrac{1}{12}\right)}=\dfrac{16}{11}$$

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