( 1 3 + 4 7 ) + ( 1 3 2 + 1 3 × 4 7 + 4 2 7 2 ) + ( 1 3 3 + 1 3 2 × 4 7 + 1 3 × 4 2 7 2 + 4 3 7 3 ) + ? upto infinite terms, is equal to [2026]

Question

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

Smart Achievers · Accepted Answer

(4)

$Let a = \frac{4}{7}, b = \frac{1}{3}$

${Multiply N}^{r} {and D}^{r} by (a - b) = \frac{4}{7} - \frac{1}{3} = \frac{5}{21}$

$\frac{1}{a - b} [(a^{2} - b^{2}) + (a^{3} - b^{3}) + (a^{4} - b^{4}) + \dots \infty]$

$= \frac{1}{a - b} [\frac{a^{2}}{1 - a} - \frac{b^{2}}{1 - b}]$ $= \frac{21}{5} [\frac{\frac{16}{49}}{1 - \frac{4}{7}} - \frac{\frac{1}{9}}{1 - \frac{1}{3}}]$

$= \frac{21}{5} [\frac{16}{21} - \frac{1}{6}]$ $= \frac{21}{5} [\frac{96 - 21}{21.6}]$

$= \frac{75}{5 \cdot 6}$ $= \frac{15}{6} = \frac{5}{2}$

Solution

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

1 $\frac{6}{5}$

2 $\frac{4}{3}$

3 $\frac{7}{4}$

4 $\frac{5}{2}$

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Solution

Q. (13+47)+(132+13×47+4272)+(133+132×47+13×4272+4373)+⋯ upto infinite terms, is equal to [2026]

1 65

2 43

3 74

4 52