`(1 2/3)^2 + (2 1/3)^2 + 3^2 + (3 2/3)^2` + ….to 10 terms , the sum is :

Options

A. `1390/9`
B. `1790/9`
C. `1990/9`
D. `2290/9`

Answer

To solve the problem `(1 2/3)^2 + (2 1/3)^2 + 3^2 + (3 2/3)^2 + ...` up to 10 terms, we can follow these steps:

### Step 1: Convert Mixed Numbers to Improper Fractions
First, we convert the mixed numbers into improper fractions:
- \(1 \frac{2}{3} = \frac{5}{3}\)
- \(2 \frac{1}{3} = \frac{7}{3}\)
- \(3 = \frac{9}{3}\)
- \(3 \frac{2}{3} = \frac{11}{3}\)

Thus, the series can be rewritten as:
\[ \left(\frac{5}{3}\right)^2 + \left(\frac{7}{3}\right)^2 + \left(\frac{9}{3}\right)^2 + \left(\frac{11}{3}\right)^2 + \ldots \]

### Step 2: Identify the General Term
The numerators of the fractions form an arithmetic progression (AP): \(5, 7, 9, 11, \ldots\). The general term \(T_n\) can be expressed as:
\[ T_n = \left(5 + (n-1) \cdot 2\right)^2 = (3 + 2n)^2 \]

### Step 3: Write the Series in Summation Form
We can express the sum of the first 10 terms as:
\[ S = \sum_{n=1}^{10} T_n = \sum_{n=1}^{10} (3 + 2n)^2 \]

### Step 4: Expand the Square
Expanding the square gives:
\[ (3 + 2n)^2 = 9 + 12n + 4n^2 \]
Thus, the sum becomes:
\[ S = \sum_{n=1}^{10} (9 + 12n + 4n^2) = \sum_{n=1}^{10} 9 + \sum_{n=1}^{10} 12n + \sum_{n=1}^{10} 4n^2 \]

### Step 5: Calculate Each Component of the Sum
1. **Sum of constants**:
\[ \sum_{n=1}^{10} 9 = 9 \times 10 = 90 \]

2. **Sum of first 10 natural numbers**:
\[ \sum_{n=1}^{10} n = \frac{10 \times 11}{2} = 55 \]
Therefore,
\[ \sum_{n=1}^{10} 12n = 12 \times 55 = 660 \]

3. **Sum of squares of the first 10 natural numbers**:
\[ \sum_{n=1}^{10} n^2 = \frac{10 \times 11 \times 21}{6} = 385 \]
Therefore,
\[ \sum_{n=1}^{10} 4n^2 = 4 \times 385 = 1540 \]

### Step 6: Combine All Parts
Now, we can combine all parts to find \(S\):
\[ S = 90 + 660 + 1540 = 2290 \]

### Step 7: Adjust for the Common Factor
Since we factored out \(\frac{1}{9}\) in the beginning, we need to multiply \(S\) by \(\frac{1}{9}\):
\[ \text{Final Sum} = \frac{2290}{9} \]

### Final Answer
Thus, the sum of the series up to 10 terms is:
\[ \frac{2290}{9} \]