The sum of the infinite series, `1 ^(2) -(2^(2))/(5) + (3 ^(2))/(5 ^(2))+ (4^(2))/(5 ^(3))+ (5 ^(2))/(5 ^(4))-(6 ^(2))/(5 ^(5)) + .....` is:

Options

A. `1/2`
B. `25/24`
C. `25/54`
D. `125/252`

Answer

To find the sum of the infinite series

\[ S = 1^2 - \frac{2^2}{5} + \frac{3^2}{5^2} + \frac{4^2}{5^3} + \frac{5^2}{5^4} - \frac{6^2}{5^5} + \ldots \]

we can analyze the series step by step.

### Step 1: Rewrite the series
We can express the series in a more manageable form. Notice that the series alternates in sign and has powers of \(5\) in the denominator. We can separate the terms based on their signs:

\[ S = 1^2 + \left( -\frac{2^2}{5} + \frac{3^2}{5^2} + \frac{4^2}{5^3} + \frac{5^2}{5^4} - \frac{6^2}{5^5} + \ldots \right) \]

### Step 2: Factor out common terms
We can factor out \(\frac{1}{5^n}\) from the series where \(n\) is the index of summation:

\[ S = 1 + \sum_{n=1}^{\infty} \frac{(-1)^{n+1} n^2}{5^n} \]

### Step 3: Recognize the series as a known form
The series \(\sum_{n=1}^{\infty} \frac{n^2 x^n}{n!}\) can be evaluated using the formula for the sum of a power series. In our case, we have \(x = -\frac{1}{5}\).

### Step 4: Use the formula for the sum of the series
The sum of the series can be calculated using the formula:

\[ \sum_{n=0}^{\infty} n^2 x^n = \frac{x(1+x)}{(1-x)^3} \]

Substituting \(x = -\frac{1}{5}\):

\[ \sum_{n=1}^{\infty} n^2 \left(-\frac{1}{5}\right)^n = \frac{-\frac{1}{5}(1 - \frac{1}{5})}{(1 + \frac{1}{5})^3} \]

### Step 5: Simplify the expression
Calculating the right-hand side:

\[ = \frac{-\frac{1}{5} \cdot \frac{4}{5}}{\left(\frac{6}{5}\right)^3} = \frac{-\frac{4}{25}}{\frac{216}{125}} = -\frac{4 \cdot 125}{25 \cdot 216} = -\frac{500}{5400} = -\frac{5}{54} \]

### Step 6: Combine results
Now, we combine this result with the first term \(1\):

\[ S = 1 - \frac{5}{54} \]

### Step 7: Final calculation
To combine these fractions:

\[ S = \frac{54}{54} - \frac{5}{54} = \frac{49}{54} \]

Thus, the sum of the infinite series is:

\[ \boxed{\frac{49}{54}} \]