The sum of the infinite series is 1/5(1/5-1/7)+(1/5)^2((1/5)^2-(1/7)^2)+(1/5)^3((1/5)^3-(1/7)^3)+ .. equal to Option 1: 5/408 Option 2: 7/816 Option 3: 7/408 Option 4: 5/816

Answer: 5/408. Solution: 1/5(1/5-1/7)+(1/5)^2((1/5)^2-(1/7)^2)+(1/5)^3((1/5)^3-(1/7)^3)+ Each term follows the pattern: (1/5)^n [(1/5)^n - (1/7)^n] where n = 1, 2, 3, Expanding each term: (1/5)^n [(1/5)^n - (1/7)^n] = (1/5)^n * (1/5)^n - (1/5)^n * (1/7)^n Using the power rule x^a * x^b = x^(a+b): = (1/5)^2n - (1/5 * 1/7)^n Simplifying: = (1/5)^2n - (1/35)^n Our series becomes: _(n=1)^(∞) [(1/5)^2n - (1/35)^n] = _(n=1)^(∞) (1/5)^2n - _(n=1)^(∞) (1/35)^n For the first series: (1/5)^2n = ((1/5)^2)^n = (1/25)^n So we have: _(n=1)^(∞) (1/25)^n - _(n=1)^(∞) (1/35)^n For geometric series with, r, The sum is: 1/24 - 1/34 = (34-24)/(24 x 34) = 10/816 = 5/408

Algebra

Progression & Series

Medium

The sum of the infinite series is $\frac{1}{5}\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{5}\right)^{2}\left(\left(\frac{1}{5}\right)^{2}-\left(\frac{1}{7}\right)^{2}\right)+\left(\frac{1}{5}\right)^{3}\left(\left(\frac{1}{5}\right)^{3}-\left(\frac{1}{7}\right)^{3}\right)+\ldots .$ . equal to

$\frac{5}{408}$

$\frac{7}{816}$

$\frac{7}{408}$

$\frac{5}{816}$

✅ Correct Option: 1

$\dfrac{1}{5}\left(\dfrac{1}{5}-\dfrac{1}{7}\right)+\left(\dfrac{1}{5}\right)^{2}\left(\left(\dfrac{1}{5}\right)^{2}-\left(\dfrac{1}{7}\right)^{2}\right)+\left(\dfrac{1}{5}\right)^{3}\left(\left(\dfrac{1}{5}\right)^{3}-\left(\dfrac{1}{7}\right)^{3}\right)+\ldots$

Each term follows the pattern:

$\left(\dfrac{1}{5}\right)^n \left[\left(\dfrac{1}{5}\right)^n - \left(\dfrac{1}{7}\right)^n\right]$

where $n = 1, 2, 3, \ldots$

Expanding each term:

$\left(\dfrac{1}{5}\right)^n \left[\left(\dfrac{1}{5}\right)^n - \left(\dfrac{1}{7}\right)^n\right] = \left(\dfrac{1}{5}\right)^n \cdot \left(\dfrac{1}{5}\right)^n - \left(\dfrac{1}{5}\right)^n \cdot \left(\dfrac{1}{7}\right)^n$

Using the power rule $x^a \cdot x^b = x^{a+b}$ :

$= \left(\dfrac{1}{5}\right)^{2n} - \left(\dfrac{1}{5} \cdot \dfrac{1}{7}\right)^n$

Simplifying:

$= \left(\dfrac{1}{5}\right)^{2n} - \left(\dfrac{1}{35}\right)^n$

Our series becomes:

$\sum_{n=1}^{\infty} \left[\left(\dfrac{1}{5}\right)^{2n} - \left(\dfrac{1}{35}\right)^n\right] = \sum_{n=1}^{\infty} \left(\dfrac{1}{5}\right)^{2n} - \sum_{n=1}^{\infty} \left(\dfrac{1}{35}\right)^n$

For the first series: $\left(\dfrac{1}{5}\right)^{2n} = \left(\left(\dfrac{1}{5}\right)^2\right)^n = \left(\dfrac{1}{25}\right)^n$

So we have:

$\sum_{n=1}^{\infty} \left(\dfrac{1}{25}\right)^n - \sum_{n=1}^{\infty} \left(\dfrac{1}{35}\right)^n$

For geometric series with $|r| < 1$ : $\sum_{n=1}^{\infty} r^n = \dfrac{r}{1-r}$

First series with $r = \dfrac{1}{25}$ :

$\sum_{n=1}^{\infty} \left(\dfrac{1}{25}\right)^n = \dfrac{\dfrac{1}{25}}{1-\dfrac{1}{25}} = \dfrac{\dfrac{1}{25}}{\dfrac{24}{25}} = \dfrac{1}{25} \times \dfrac{25}{24} = \dfrac{1}{24}$

Second series with $r = \dfrac{1}{35}$ :

$\sum_{n=1}^{\infty} \left(\dfrac{1}{35}\right)^n = \dfrac{\dfrac{1}{35}}{1-\dfrac{1}{35}} = \dfrac{\dfrac{1}{35}}{\dfrac{34}{35}} = \dfrac{1}{35} \times \dfrac{35}{34} = \dfrac{1}{34}$

The sum is:

$\dfrac{1}{24} - \dfrac{1}{34} = \dfrac{34-24}{24 \times 34} = \dfrac{10}{816} = \dfrac{5}{408}$

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