The sum `(3 xx 1^(3))/(1^(2)) + (5 xx (1^(3) + 2^(3)))/(1^(2) + 2^(2)) + (7 xx (1^(3) + 2^(3) + 3^(3)))/(1^(2) +2^(2) + 3^(2))...` upto 10th term is

Options

A. 680
B. 620
C. 660
D. 600

Explanation

To solve the given series, we need to find the sum of the series up to the 10th term:

\[ S = \sum_{n=1}^{10} \frac{(2n + 1) \cdot \sum_{k=1}^{n} k^3}{\sum_{k=1}^{n} k^2} \]

### Step 1: Identify the general term

The general term of the series can be expressed as:

\[ T_n = \frac{(2n + 1) \cdot \sum_{k=1}^{n} k^3}{\sum_{k=1}^{n} k^2} \]

### Step 2: Use formulas for the sums

We know the formulas for the sums of cubes and squares:

1. The sum of the first \( n \) cubes is given by:

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

2. The sum of the first \( n \) squares is given by:

\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 3: Substitute the formulas into the general term

Substituting these formulas into \( T_n \):

\[ T_n = \frac{(2n + 1) \cdot \left( \frac{n(n + 1)}{2} \right)^2}{\frac{n(n + 1)(2n + 1)}{6}} \]

### Step 4: Simplify the expression

Now, simplify \( T_n \):

\[ T_n = \frac{(2n + 1) \cdot \frac{n^2(n + 1)^2}{4}}{\frac{n(n + 1)(2n + 1)}{6}} \]

Cancelling out \( (2n + 1) \), \( n \), and \( n + 1 \):

\[ T_n = \frac{6n(n + 1)}{4} = \frac{3n(n + 1)}{2} \]

### Step 5: Find the total sum

Now, we need to find the sum \( S \):

\[ S = \sum_{n=1}^{10} T_n = \sum_{n=1}^{10} \frac{3n(n + 1)}{2} \]

### Step 6: Split the summation

This can be split into two separate summations:

\[ S = \frac{3}{2} \sum_{n=1}^{10} n^2 + \frac{3}{2} \sum_{n=1}^{10} n \]

### Step 7: Use formulas for summations

Using the formulas for the sums:

1. The sum of the first \( n \) natural numbers:

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

2. The sum of the squares of the first \( n \) natural numbers:

\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 8: Substitute \( n = 10 \)

Calculating these for \( n = 10 \):

\[ \sum_{k=1}^{10} k = \frac{10 \cdot 11}{2} = 55 \]

\[ \sum_{k=1}^{10} k^2 = \frac{10 \cdot 11 \cdot 21}{6} = 385 \]

### Step 9: Substitute back into the sum

Now substituting back into \( S \):

\[ S = \frac{3}{2} \cdot 385 + \frac{3}{2} \cdot 55 = \frac{3}{2} (385 + 55) = \frac{3}{2} \cdot 440 = 660 \]

### Final Answer

Thus, the sum of the series up to the 10th term is:

\[ \boxed{660} \]