If `S_n` denotes the sum of first n natural number then `S_1 + S_2x + S_3 x^2 +……+S_n x^(n-1) + ……oo` terms =

Options

A. `(1 -x )^(-1)`
B. `(1-x)^(-2)`
C. `(1 -x)^(-3)`
D. `(1 - x)^(-4)`

Answer

To solve the problem, we need to find the sum of the series given by:

\[ S_1 + S_2x + S_3x^2 + \ldots + S_n x^{n-1} + \ldots \]

where \( S_n \) is the sum of the first \( n \) natural numbers. The formula for \( S_n \) is:

\[ S_n = \frac{n(n + 1)}{2} \]

### Step 1: Express the series using the formula for \( S_n \)

Substituting \( S_n \) into the series, we get:

\[ \sum_{n=1}^{\infty} S_n x^{n-1} = \sum_{n=1}^{\infty} \frac{n(n + 1)}{2} x^{n-1} \]

### Step 2: Factor out the constant

We can factor out \( \frac{1}{2} \):

\[ = \frac{1}{2} \sum_{n=1}^{\infty} n(n + 1) x^{n-1} \]

### Step 3: Separate the summation

Now, we can separate the summation into two parts:

\[ = \frac{1}{2} \left( \sum_{n=1}^{\infty} n^2 x^{n-1} + \sum_{n=1}^{\infty} n x^{n-1} \right) \]

### Step 4: Use known summation formulas

We know the following formulas for the sums:

1. The sum of \( n x^{n-1} \) is given by:

\[ \sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1 - x)^2} \quad \text{for } |x| < 1 \]

2. The sum of \( n^2 x^{n-1} \) can be derived using the first formula:

\[ \sum_{n=1}^{\infty} n^2 x^{n-1} = x \frac{d}{dx} \left( \frac{1}{(1 - x)^2} \right) = \frac{x(1 + x)}{(1 - x)^3} \]

### Step 5: Substitute the summation results

Now substituting these results back into our equation:

\[ = \frac{1}{2} \left( \frac{x(1 + x)}{(1 - x)^3} + \frac{1}{(1 - x)^2} \right) \]

### Step 6: Combine the fractions

To combine these fractions, we need a common denominator:

\[ = \frac{1}{2} \left( \frac{x(1 + x) + (1 - x)}{(1 - x)^3} \right) \]

### Step 7: Simplify the numerator

Now simplify the numerator:

\[ = \frac{1}{2} \left( \frac{x + x^2 + 1 - x}{(1 - x)^3} \right) = \frac{1}{2} \left( \frac{x^2 + 1}{(1 - x)^3} \right) \]

### Step 8: Final result

Thus, the final result for the series is:

\[ = \frac{x^2 + 1}{2(1 - x)^3} \]

### Conclusion

The sum of the series \( S_1 + S_2x + S_3x^2 + \ldots \) is:

\[ \frac{x^2 + 1}{2(1 - x)^3} \]