Sum of the n terms of the series `(3)/(1^(2))+(5)/(1^(2)+2^(2))+(7)/(1^(2)+2^(2)+3^(3))+"......."` is

Options

A. `(2n)/(n+1)`
B. `(4n)/(n+1)`
C. `(6n)/(n+1)`
D. `(9n)/(n+1)`

Answer

To find the sum of the n terms of the series

\[ S_n = \frac{3}{1^2} + \frac{5}{1^2 + 2^2} + \frac{7}{1^2 + 2^2 + 3^2} + \ldots \]

we can break down the problem step by step.

### Step 1: Identify the Numerator
The numerators of the terms in the series are:
- First term: 3
- Second term: 5
- Third term: 7

We can see that the numerators form an arithmetic progression (AP) with the first term \(a = 3\) and a common difference \(d = 2\).

The \(n\)-th term of the numerator can be expressed as:
\[ \text{Numerator} = 3 + (n-1) \cdot 2 = 2n + 1 \]

### Step 2: Identify the Denominator
The denominators of the terms are:
- First term: \(1^2\)
- Second term: \(1^2 + 2^2\)
- Third term: \(1^2 + 2^2 + 3^2\)

The denominator for the \(n\)-th term is the sum of squares of the first \(n\) natural numbers, which can be expressed as:
\[ \text{Denominator} = 1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n+1)(2n+1)}{6} \]

### Step 3: Combine the Terms
Thus, the \(n\)-th term of the series can be expressed as:
\[ T_n = \frac{2n + 1}{\frac{n(n+1)(2n+1)}{6}} = \frac{6(2n + 1)}{n(n + 1)(2n + 1)} \]

### Step 4: Simplify the Expression
We can simplify this expression:
\[ T_n = \frac{6}{n(n + 1)} \]

### Step 5: Find the Sum of the Series
Now, we need to find the sum of the first \(n\) terms:
\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{6}{k(k + 1)} \]

We can use partial fractions to simplify \(\frac{6}{k(k + 1)}\):
\[ \frac{6}{k(k + 1)} = 6 \left(\frac{1}{k} - \frac{1}{k + 1}\right) \]

### Step 6: Write the Sum
Thus, the sum becomes:
\[ S_n = 6 \sum_{k=1}^{n} \left(\frac{1}{k} - \frac{1}{k + 1}\right) \]

### Step 7: Evaluate the Sum
This is a telescoping series, where most terms cancel out:
\[ S_n = 6 \left(1 - \frac{1}{n + 1}\right) = 6 \left(\frac{n}{n + 1}\right) \]

### Final Result
Thus, the sum of the first \(n\) terms of the series is:
\[ S_n = \frac{6n}{n + 1} \]