Question ID: #886
The sum of the series $(\frac{1}{3}+\frac{4}{7})+(\frac{1}{3^{2}}+\frac{1}{3}\times\frac{4}{7}+\frac{4^{2}}{7^{2}})+(\frac{1}{3^{3}}+\frac{1}{3^{2}}\times\frac{4}{7}+\frac{1}{3}\times\frac{4^{2}}{7^{2}}+\frac{4^{3}}{7^{3}})+…..$ upto infinite terms is equal to
- (1) $\frac{5}{2}$
- (2) $\frac{7}{4}$
- (3) $\frac{4}{3}$
- (4) $\frac{6}{5}$
Solution:
Let $x = \frac{1}{3}$ and $y = \frac{4}{7}$.
The series is $(x+y) + (x^2+xy+y^2) + (x^3+x^2y+xy^2+y^3) + \dots$
Multiply and divide by $(x-y)$:
$$S = \frac{1}{x-y} \left[ (x^2-y^2) + (x^3-y^3) + (x^4-y^4) + \dots \right]$$
$$S = \frac{1}{x-y} \left[ (x^2+x^3+x^4+\dots) – (y^2+y^3+y^4+\dots) \right]$$
Using sum of infinite GP formula $\frac{a}{1-r}$:
$$S = \frac{1}{x-y} \left[ \frac{x^2}{1-x} – \frac{y^2}{1-y} \right]$$
Substitute values $x = \frac{1}{3}, y = \frac{4}{7}$:
$$x-y = \frac{1}{3} – \frac{4}{7} = \frac{7-12}{21} = -\frac{5}{21}$$
$$\frac{x^2}{1-x} = \frac{1/9}{2/3} = \frac{1}{6}$$
$$\frac{y^2}{1-y} = \frac{16/49}{3/7} = \frac{16}{21}$$
$$S = \frac{1}{-5/21} \left[ \frac{1}{6} – \frac{16}{21} \right] = -\frac{21}{5} \left[ \frac{7 – 32}{42} \right]$$
$$S = -\frac{21}{5} \left[ \frac{-25}{42} \right] = \frac{21}{5} \cdot \frac{25}{42} = \frac{1}{1} \cdot \frac{5}{2} = \frac{5}{2}$$
Ans. (1)
Was this solution helpful?
YesNo