The sum of the series `1+(9)/(4)+(36)/(9)+(100)/(16)+…` infinite terms is

Options

A. `446`
B. `746`
C. `546`
D. `846`

Explanation

To find the sum of the infinite series \( S = 1 + \frac{9}{4} + \frac{36}{9} + \frac{100}{16} + \ldots \), we will first identify the pattern in the series and then derive a formula for the \( n \)-th term.

### Step 1: Identify the pattern
The terms of the series can be rewritten as:
- The first term is \( 1 = \frac{1^2}{1^2} \)
- The second term is \( \frac{9}{4} = \frac{3^2}{2^2} \)
- The third term is \( \frac{36}{9} = \frac{6^2}{3^2} \)
- The fourth term is \( \frac{100}{16} = \frac{10^2}{4^2} \)

From this, we can observe that the \( n \)-th term can be expressed as:
\[ T_n = \frac{(n(n+1))^2}{(n^2)^2} = \frac{(n(n+1))^2}{n^2} \]

### Step 2: Simplify the \( n \)-th term
We can simplify the \( n \)-th term:
\[ T_n = \frac{n^2(n+1)^2}{n^2} = (n+1)^2 \]

### Step 3: Write the series in summation form
The series can now be expressed as:
\[ S = \sum_{n=1}^{\infty} (n+1)^2 \]

### Step 4: Find the sum of the series
To find the sum of the series, we can use the formula for the sum of squares:
\[ \sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6} \]
However, since we are looking for an infinite series, we need to recognize that the series diverges.

### Conclusion
The series diverges as \( n \) approaches infinity, meaning it does not converge to a finite sum.