Value of `(1^2)/1+(1^2+2^2)/(1+2)+(1^2+2^2+3^2)/(1+2+3)+.... 'n'` terms is equal to

Options

A. `(2n^(2))/(n+1)`
B. `(n(n+2))/(3)`
C. `(4n^(2))/(3n+2)`
D. none of these

Answer

To solve the problem, we need to evaluate the expression:

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

### Step 1: Identify the general term

The general term in the series can be expressed as:

\[ T_r = \frac{1^2 + 2^2 + \ldots + r^2}{1 + 2 + \ldots + r} \]

### Step 2: Use formulas for sums

We know the formulas for the sum of the first \( r \) squares and the sum of the first \( r \) natural numbers:

1. The sum of the first \( r \) squares:
\[ 1^2 + 2^2 + \ldots + r^2 = \frac{r(r + 1)(2r + 1)}{6} \]

2. The sum of the first \( r \) natural numbers:
\[ 1 + 2 + \ldots + r = \frac{r(r + 1)}{2} \]

### Step 3: Substitute the formulas into \( T_r \)

Substituting these formulas into \( T_r \):

\[ T_r = \frac{\frac{r(r + 1)(2r + 1)}{6}}{\frac{r(r + 1)}{2}} \]

### Step 4: Simplify \( T_r \)

Now, we simplify \( T_r \):

\[ T_r = \frac{r(r + 1)(2r + 1)}{6} \cdot \frac{2}{r(r + 1)} = \frac{2r + 1}{3} \]

### Step 5: Sum up all terms

Now, we need to sum \( T_r \) from \( r = 1 \) to \( n \):

\[ S_n = \sum_{r=1}^{n} T_r = \sum_{r=1}^{n} \frac{2r + 1}{3} \]

### Step 6: Factor out the constant

Factoring out \( \frac{1}{3} \):

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

### Step 7: Calculate the sums

We know:

\[ \sum_{r=1}^{n} r = \frac{n(n + 1)}{2} \]
and
\[ \sum_{r=1}^{n} 1 = n \]

Substituting these into the equation:

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

### Step 8: Simplify further

This simplifies to:

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

### Step 9: Final expression

Thus, we can write:

\[ S_n = \frac{n(n + 2)}{3} \]

### Final Answer:

The value of the given expression for \( n \) terms is:

\[ \boxed{\frac{n(n + 2)}{3}} \]