upto infinite terms, is equal to: [2026]

Question

$(\frac{1}{3} + \frac{4}{7}) + (\frac{1}{3^{2}} + \frac{1}{3} \times \frac{4}{7} + \frac{4^{2}}{7^{2}}) + (\frac{1}{3^{2}} + \frac{1}{3^{2}} \times \frac{4}{7} + \frac{1}{3} \times \frac{4^{2}}{7^{2}} + \frac{4^{3}}{7^{3}}) + . . .$ upto infinite terms, is equal to: [2026]

Smart Achievers · Accepted Answer

(2)

Let $S = (\frac{1}{3} + \frac{4}{7}) + (\frac{1}{3^{2}} + \frac{1}{3} \times \frac{4}{7} + \frac{4^{2}}{7^{2}}) + . . . \infty$

Multiply both sides by $(\frac{1}{3} - \frac{4}{7})$

$S (\frac{1}{3} - \frac{4}{7}) = {(\frac{1}{3})}^{2} - {(\frac{4}{7})}^{2} + {(\frac{1}{3})}^{3} - {(\frac{4}{7})}^{3} + {(\frac{1}{3})}^{4} - {(\frac{4}{7})}^{4} + . . . \infty$

$\Rightarrow S (\frac{- 5}{21}) = ({(\frac{1}{3})}^{2} + {(\frac{1}{3})}^{3} + . . . \infty) - ({(\frac{4}{7})}^{2} + {(\frac{4}{7})}^{3} + . . . \infty)$

$= \frac{{(\frac{1}{3})}^{2}}{1 - \frac{1}{3}} - \frac{{(\frac{4}{7})}^{2}}{1 - \frac{4}{7}} = \frac{1}{9} \times \frac{3}{2} - \frac{16}{49} \times \frac{7}{3}$

$= \frac{1}{6} - \frac{16}{21} = \frac{7 - 32}{42}$

$\Rightarrow S (\frac{- 5}{21}) = \frac{- 25}{42} \Rightarrow S = \frac{5}{2}$

Solution

Q.
$(\frac{1}{3} + \frac{4}{7}) + (\frac{1}{3^{2}} + \frac{1}{3} \times \frac{4}{7} + \frac{4^{2}}{7^{2}}) + (\frac{1}{3^{2}} + \frac{1}{3^{2}} \times \frac{4}{7} + \frac{1}{3} \times \frac{4^{2}}{7^{2}} + \frac{4^{3}}{7^{3}}) + . . .$ upto infinite terms, is equal to: [2026]

1 $\frac{4}{3}$

2 $\frac{5}{2}$

3 $\frac{6}{5}$

4 $\frac{7}{4}$

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Solution

Q. (13+47)+(132+13×47+4272)+(132+132×47+13×4272+4373)+... upto infinite terms, is equal to: [2026]

1 43

2 52

3 65

4 74