The sum to infinity of the series
`1+2(1-(1)/(n))+3(1-(1)/(n))^(2)+ . . .. . . .`, is (A)`n^(2)` (B)`n(n+1)` (C)`n(1+(1)/(n))^(2)` (D)None of these

Options

A. `n^(2)`
B. `n(n+1)`
C. `n(1+(1)/(n))^(2)`
D. None of these

Explanation

To find the sum to infinity of the series given by:

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

we can follow these steps:

### Step 1: Define the series
We start with the series:

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

### Step 2: Multiply the series by \( \left(1 - \frac{1}{n}\right) \)
Now, we multiply \( S_n \) by \( \left(1 - \frac{1}{n}\right) \):

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

### Step 3: Subtract the two series
Next, we subtract the second series from the first:

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

This simplifies to:

\[ S_n - S_n \left(1 - \frac{1}{n}\right) = 1 + \left(1 - \frac{1}{n}\right) + \left(1 - \frac{1}{n}\right)^2 + \ldots \]

### Step 4: Recognize the right-hand side as a geometric series
The right-hand side is a geometric series with the first term \( a = 1 \) and common ratio \( r = \left(1 - \frac{1}{n}\right) \):

The sum of an infinite geometric series is given by:

\[ \text{Sum} = \frac{a}{1 - r} \]

So we have:

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

### Step 5: Substitute back into the equation
Now, substituting back into our equation:

\[ S_n - S_n \left(1 - \frac{1}{n}\right) = n \]

### Step 6: Factor out \( S_n \)
This gives us:

\[ S_n \left(1 - \left(1 - \frac{1}{n}\right)\right) = n \]

This simplifies to:

\[ S_n \cdot \frac{1}{n} = n \]

### Step 7: Solve for \( S_n \)
Multiplying both sides by \( n \):

\[ S_n = n^2 \]

### Conclusion
Thus, the sum to infinity of the series is:

\[ S_n = n^2 \]

The correct answer is (A) \( n^2 \).

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