Sum of series `1 + 2x + 7x^2 + 20x^3 + .......` up to n terms when `x = - 1` is

Options

A. (a) `(1)/(16) (4n + 3+ (-3)^(n+1))`
B. (b) `(1)/(16) (4n - 3+ (-3)^(n+1))`
C. (c) `(1)/(16) (4n + 3+ 3^(n+1))`
D. (d) `(1)/(16) (4n + 3 - 3^(n+1))`

Answer

To solve the problem of finding the sum of the series \(1 + 2x + 7x^2 + 20x^3 + \ldots\) up to \(n\) terms when \(x = -1\), we can follow these steps:

### Step 1: Identify the series
The series given is:
\[ S_n = 1 + 2x + 7x^2 + 20x^3 + \ldots \]

### Step 2: Substitute \(x = -1\)
We need to evaluate the series when \(x = -1\):
\[ S_n = 1 + 2(-1) + 7(-1)^2 + 20(-1)^3 + \ldots \]
This simplifies to:
\[ S_n = 1 - 2 + 7 - 20 + \ldots \]

### Step 3: Calculate the first few sums
Let's calculate the sums for the first few terms:

- For \(n = 1\):
\[ S_1 = 1 \]

- For \(n = 2\):
\[ S_2 = 1 + 2(-1) = 1 - 2 = -1 \]

- For \(n = 3\):
\[ S_3 = 1 + 2(-1) + 7(-1)^2 = 1 - 2 + 7 = 6 \]

- For \(n = 4\):
\[ S_4 = 1 + 2(-1) + 7(-1)^2 + 20(-1)^3 = 1 - 2 + 7 - 20 = -14 \]

### Step 4: Identify the pattern
From the calculations:
- \(S_1 = 1\)
- \(S_2 = -1\)
- \(S_3 = 6\)
- \(S_4 = -14\)

### Step 5: Check the options
Now we need to check the provided options to see which one matches these sums.

**Option A:**
\[ \frac{1}{16}(4n + 3 + (-3)^{n+1}) \]
Calculating for \(n = 1\):
\[ S_1 = \frac{1}{16}(4(1) + 3 + (-3)^{2}) = \frac{1}{16}(4 + 3 + 9) = \frac{16}{16} = 1 \]

Calculating for \(n = 2\):
\[ S_2 = \frac{1}{16}(4(2) + 3 + (-3)^{3}) = \frac{1}{16}(8 + 3 - 27) = \frac{1}{16}(-16) = -1 \]

Calculating for \(n = 3\):
\[ S_3 = \frac{1}{16}(4(3) + 3 + (-3)^{4}) = \frac{1}{16}(12 + 3 + 81) = \frac{1}{16}(96) = 6 \]

Calculating for \(n = 4\):
\[ S_4 = \frac{1}{16}(4(4) + 3 + (-3)^{5}) = \frac{1}{16}(16 + 3 - 243) = \frac{1}{16}(-224) = -14 \]

### Conclusion
Since all calculated sums \(S_1, S_2, S_3, S_4\) match with the values derived from Option A, we conclude that the correct answer is:
\[ \text{Option A: } \frac{1}{16}(4n + 3 + (-3)^{n+1}) \]