The value of 1+ ≤ft(1+1/3) \; 1/4+ ≤ft(1+1/3+1/9) \; 1/16+ ≤ft(1+1/3+1/9+1/27) \; 1/64+----- is

Correct Answer is : 161116/111116

Correct Answer is : 161116/111116

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CAT 2023 Slot 3 Solved Paper

Question : 64 of 66

Marks: +1, -0

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}+-----

Solution:

The given sequence can be written as:

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

We know that the sum of an infinite G.P. is

\frac{a}{1-r}

, where

a

is the first term and

r

is the common ratio.

\Rightarrow

The first term

=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}

⇒ The second term

=\frac{1}{3} \left(\frac{\frac{1}{4}}{1-\frac{1}{4}}\right) =\frac{1}{9}

⇒ The third term

=\frac{1}{9} \left(\frac{\frac{1}{16}}{1-\frac{1}{4}}\right) =\frac{1}{108}

Observing these three terms, we see that they are in G.P. with a common ratio of

\frac{1}{12}

⇒ Sum of this infinite G.P.

=\frac{\frac{4}{3}}{1-\frac{1}{12}} =\frac{16}{11}

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