If `S=2+6/7+12/7^2+20/7^3+. . . ` then find `4S`

Options

A. `(7/2)^2`
B. `(7/3)^3`
C. `(7/3)`
D. `(7/3)^4`

Answer

To solve the series \( S = 2 + \frac{6}{7} + \frac{12}{7^2} + \frac{20}{7^3} + \ldots \), we can identify a pattern in the terms. The numerators appear to be following a specific sequence, while the denominators are powers of 7.

### Step 1: Identify the pattern in the series
The series can be expressed as:
\[ S = 2 + \frac{6}{7} + \frac{12}{7^2} + \frac{20}{7^3} + \ldots \]
The numerators are \( 2, 6, 12, 20, \ldots \). Observing the differences:
- \( 6 - 2 = 4 \)
- \( 12 - 6 = 6 \)
- \( 20 - 12 = 8 \)

The differences \( 4, 6, 8, \ldots \) suggest that the numerators can be expressed in terms of \( n(n+1) \) where \( n \) is the term index starting from 1.

### Step 2: Express the numerators
The numerators can be rewritten as:
- For \( n=1 \): \( 2 = 1 \cdot 2 \)
- For \( n=2 \): \( 6 = 2 \cdot 3 \)
- For \( n=3 \): \( 12 = 3 \cdot 4 \)
- For \( n=4 \): \( 20 = 4 \cdot 5 \)

Thus, we can express the \( n \)-th term as:
\[ \frac{n(n+1)}{7^{n-1}} \]

### Step 3: Rewrite the series
The series can now be rewritten as:
\[ S = \sum_{n=1}^{\infty} \frac{n(n+1)}{7^{n-1}} \]

### Step 4: Use the formula for power series
We can use the formula for the sum of the series:
\[ \sum_{n=0}^{\infty} n x^n = \frac{x}{(1-x)^2} \]
and
\[ \sum_{n=0}^{\infty} n^2 x^n = x \frac{1+x}{(1-x)^3} \]

### Step 5: Calculate \( S \)
Using \( x = \frac{1}{7} \):
\[ \sum_{n=1}^{\infty} n x^n = \frac{\frac{1}{7}}{(1 - \frac{1}{7})^2} = \frac{\frac{1}{7}}{\left(\frac{6}{7}\right)^2} = \frac{1}{7} \cdot \frac{49}{36} = \frac{7}{36} \]
And for \( n(n+1) \):
\[ \sum_{n=1}^{\infty} n(n+1)x^n = x \frac{1+x}{(1-x)^3} = \frac{1/7 \cdot (1 + \frac{1}{7})}{(1 - \frac{1}{7})^3} \]
Calculating this gives:
\[ = \frac{\frac{1}{7} \cdot \frac{8}{7}}{\left(\frac{6}{7}\right)^3} = \frac{\frac{8}{49}}{\frac{216}{343}} = \frac{8 \cdot 343}{49 \cdot 216} = \frac{8 \cdot 7}{216} = \frac{56}{216} = \frac{7}{27} \]

### Step 6: Find \( 4S \)
Now, we multiply \( S \) by 4:
\[ 4S = 4 \cdot \frac{7}{27} = \frac{28}{27} \]

### Final Answer
Thus, the final answer is:
\[ 4S = \frac{28}{27} \]