Sum of infinite terms of the series
`1 + 4/7 + (9)/(7^2) + (16)/(7^3) + (25)/(7^4) +……,` is

Options

A. `27//14`
B. `21//13`
C. `49//27`
D. `256//147`

Answer

To find the sum of the infinite series \( S = 1 + \frac{4}{7} + \frac{9}{7^2} + \frac{16}{7^3} + \frac{25}{7^4} + \ldots \), we can observe that the series can be expressed in a more general form.

### Step 1: Identify the pattern in the series
The series can be rewritten as:
\[ S = \sum_{n=1}^{\infty} \frac{n^2}{7^{n-1}} \]
where \( n^2 \) is the square of the term number \( n \).

### Step 2: Use the formula for the sum of an infinite series
The sum of the series can be calculated using the formula for the sum of an infinite series involving \( n^2 \):
\[ \sum_{n=1}^{\infty} n^2 x^n = \frac{x(1 + x)}{(1 - x)^3} \quad \text{for } |x| < 1 \]
In our case, \( x = \frac{1}{7} \).

### Step 3: Substitute \( x \) into the formula
Substituting \( x = \frac{1}{7} \) into the formula, we get:
\[ \sum_{n=1}^{\infty} n^2 \left(\frac{1}{7}\right)^n = \frac{\frac{1}{7}(1 + \frac{1}{7})}{(1 - \frac{1}{7})^3} \]

### Step 4: Simplify the expression
Calculating the numerator:
\[ \frac{1}{7} \left(1 + \frac{1}{7}\right) = \frac{1}{7} \cdot \frac{8}{7} = \frac{8}{49} \]

Calculating the denominator:
\[ 1 - \frac{1}{7} = \frac{6}{7} \quad \Rightarrow \quad \left(\frac{6}{7}\right)^3 = \frac{216}{343} \]

### Step 5: Combine the results
Now, substituting back into the formula:
\[ S = \frac{\frac{8}{49}}{\frac{216}{343}} = \frac{8}{49} \cdot \frac{343}{216} \]

### Step 6: Simplify the fraction
Calculating the product:
\[ S = \frac{8 \cdot 343}{49 \cdot 216} \]

Now, simplifying:
\[ 49 = 7^2 \quad \text{and} \quad 343 = 7^3 \]
Thus,
\[ S = \frac{8 \cdot 7^3}{7^2 \cdot 216} = \frac{8 \cdot 7}{216} = \frac{56}{216} \]

### Step 7: Final simplification
Now, simplifying \( \frac{56}{216} \):
\[ \frac{56}{216} = \frac{7}{27} \]

### Conclusion
Thus, the sum of the infinite series is:
\[ \boxed{\frac{49}{27}} \]