The value `5/(2^2. 3^2)+7/(3^2 .4^2)+9/(4^2. 5^2)+11/(5^2. 6^2)+13/(6^2. 7^2)+15/(7^2. 8^2)+17/(8^2. 9^2)+19/(9^2. 10^2)` is equal to

Options

A. `1/100`
B. `6/25`
C. `101/100`
D. 1

Answer

To solve the expression

\[ \frac{5}{2^2 \cdot 3^2} + \frac{7}{3^2 \cdot 4^2} + \frac{9}{4^2 \cdot 5^2} + \frac{11}{5^2 \cdot 6^2} + \frac{13}{6^2 \cdot 7^2} + \frac{15}{7^2 \cdot 8^2} + \frac{17}{8^2 \cdot 9^2} + \frac{19}{9^2 \cdot 10^2} \]

we can simplify each term using the identity:

\[ \frac{n^2 - (n-1)^2}{(n-1)^2 \cdot n^2} = \frac{2n - 1}{(n-1)^2 \cdot n^2} \]

### Step-by-Step Solution:

1. **Rewrite Each Term**:
We can express each term in the series as a difference of fractions:
\[ \frac{5}{2^2 \cdot 3^2} = \frac{9 - 4}{2^2 \cdot 3^2} = \frac{9}{2^2 \cdot 3^2} - \frac{4}{2^2 \cdot 3^2} \]
\[ \frac{7}{3^2 \cdot 4^2} = \frac{16 - 9}{3^2 \cdot 4^2} = \frac{16}{3^2 \cdot 4^2} - \frac{9}{3^2 \cdot 4^2} \]
Continuing this pattern, we can express all terms similarly.

2. **Apply the Pattern**:
Continuing this, we get:
\[ \frac{9}{3^2 \cdot 4^2} - \frac{4}{2^2 \cdot 3^2} + \frac{16}{3^2 \cdot 4^2} - \frac{9}{3^2 \cdot 4^2} + \frac{25}{4^2 \cdot 5^2} - \frac{16}{3^2 \cdot 4^2} + \ldots \]

3. **Combine Terms**:
Notice that most terms will cancel out:
\[ \left(\frac{1}{2^2} - \frac{1}{10^2}\right) \]

4. **Calculate the Remaining Terms**:
The remaining terms after cancellation will be:
\[ \frac{1}{2^2} - \frac{1}{10^2} = \frac{1}{4} - \frac{1}{100} \]

5. **Find a Common Denominator**:
The common denominator for 4 and 100 is 100:
\[ \frac{25}{100} - \frac{1}{100} = \frac{24}{100} \]

6. **Simplify the Fraction**:
Simplifying \(\frac{24}{100}\):
\[ \frac{24}{100} = \frac{6}{25} \]

### Final Answer:
Thus, the value of the expression is

\[ \frac{6}{25} \]