1. Key Concepts and Formulas

• Telescoping Series: A series where most terms cancel out when summed, typically of the form $\sum_{n=1}^{N} (f(n) - f(n+k))$. The sum simplifies to a combination of the first and last few terms.
• General Term of an Arithmetic Progression (AP): $a_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference.
• Algebraic Manipulation: Techniques like partial fraction decomposition or rewriting numerators to match components of the denominator are crucial for transforming terms into a telescoping form.

2. Step-by-Step Solution

Step 1: Identify the General Term ($T_n$) of the Series

The given series is ${3 \over {{1^2} \times {2^2}}} + {5 \over {{2^2} \times {3^2}}} + {7 \over {{3^2} \times {4^2}}} + ....$ We observe the pattern for the $n$-th term:

• Numerator: The numerators are $3, 5, 7, \dots$. This is an arithmetic progression with first term $a=3$ and common difference $d=2$. The $n$-th term is $3 + (n-1)2 = 3 + 2n - 2 = 2n+1$.
• Denominator: The denominators are $1^2 \times 2^2$, $2^2 \times 3^2$, $3^2 \times 4^2$, $\dots$. The $n$-th term is $n^2 \times (n+1)^2$.

Therefore, the general term $T_n$ is: $T_n = \frac{2n+1}{n^2 (n+1)^2}$

Step 2: Transform the General Term into a Telescoping Form

We aim to express $T_n$ in the form $f(n) - f(n+1)$. Notice that the numerator $2n+1$ can be obtained by taking the difference of the squares of the terms in the denominator: $(n+1)^2 - n^2 = (n^2 + 2n + 1) - n^2 = 2n+1$ Substitute this into the expression for $T_n$: $T_n = \frac{(n+1)^2 - n^2}{n^2 (n+1)^2}$ Now, split the fraction into two parts: $T_n = \frac{(n+1)^2}{n^2 (n+1)^2} - \frac{n^2}{n^2 (n+1)^2}$ Simplify by cancelling common factors: $T_n = \frac{1}{n^2} - \frac{1}{(n+1)^2}$ This is the telescoping form, where $f(n) = \frac{1}{n^2}$.

Step 3: Calculate the Sum of the First 10 Terms ($S_{10}$)

We need to find the sum of the first 10 terms, $S_{10} = \sum_{n=1}^{10} T_n$. $S_{10} = \sum_{n=1}^{10} \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right)$ Let's write out the terms to see the cancellation: For $n=1$: $\left(\frac{1}{1^2} - \frac{1}{2^2}\right)$ For $n=2$: $\left(\frac{1}{2^2} - \frac{1}{3^2}\right)$ For $n=3$: $\left(\frac{1}{3^2} - \frac{1}{4^2}\right)$ ... For $n=9$: $\left(\frac{1}{9^2} - \frac{1}{10^2}\right)$ For $n=10$: $\left(\frac{1}{10^2} - \frac{1}{11^2}\right)$

Summing these terms, we get: $S_{10} = \left(\frac{1}{1^2} - \cancel{\frac{1}{2^2}}\right) + \left(\cancel{\frac{1}{2^2}} - \cancel{\frac{1}{3^2}}\right) + \left(\cancel{\frac{1}{3^2}} - \cancel{\frac{1}{4^2}}\right) + \dots + \left(\cancel{\frac{1}{10^2}} - \frac{1}{11^2}\right)$ The intermediate terms cancel out, leaving: $S_{10} = \frac{1}{1^2} - \frac{1}{11^2}$ $S_{10} = 1 - \frac{1}{121}$ $S_{10} = \frac{121 - 1}{121} = \frac{120}{121}$

3. Common Mistakes & Tips

• Incorrect General Term: Double-check the pattern for the numerator and denominator carefully.
• Failure to Recognize Telescoping Form: If the denominator involves products of consecutive terms or squares of consecutive terms, look for ways to express the numerator as a difference of related terms.
• Mistakes in Cancellation: Clearly write out the terms to ensure correct identification of the terms that remain after cancellation.

4. Summary

The problem involves finding the sum of the first 10 terms of a series. By identifying the general term $T_n = \frac{2n+1}{n^2(n+1)^2}$ and cleverly rewriting the numerator as $(n+1)^2 - n^2$, we transformed $T_n$ into the telescoping form $\frac{1}{n^2} - \frac{1}{(n+1)^2}$. Summing the first 10 terms resulted in the cancellation of intermediate terms, leaving $1 - \frac{1}{11^2}$, which simplifies to $\frac{120}{121}$.

The final answer is $\boxed{\frac{120}{121}}$.