Key Concepts and Formulas

• Arithmetic Progression (AP): A sequence of numbers such that the difference between the consecutive terms is constant. The $n^{th}$ term is given by $a_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference.
• Sum of Squares of First $n$ Natural Numbers: The formula for the sum of the squares of the first $n$ natural numbers is $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.
• Algebraic Manipulation: Simplifying expressions by factoring and combining terms.

Step-by-Step Solution

1. Convert Mixed Fractions and Identify the General Term: The given series is ${\left( {1{3 \over 5}} \right)^2} + {\left( {2{2 \over 5}} \right)^2} + {\left( {3{1 \over 5}} \right)^2} + {4^2} + {\left( {4{4 \over 5}} \right)^2} + \dots$. First, convert the mixed fractions to improper fractions: $1\frac{3}{5} = \frac{5 \times 1 + 3}{5} = \frac{8}{5}$ $2\frac{2}{5} = \frac{5 \times 2 + 2}{5} = \frac{12}{5}$ $3\frac{1}{5} = \frac{5 \times 3 + 1}{5} = \frac{16}{5}$ $4 = \frac{20}{5}$ $4\frac{4}{5} = \frac{5 \times 4 + 4}{5} = \frac{24}{5}$

   The series can be rewritten as: ${\left( \frac{8}{5} \right)^2} + {\left( \frac{12}{5} \right)^2} + {\left( \frac{16}{5} \right)^2} + {\left( \frac{20}{5} \right)^2} + {\left( \frac{24}{5} \right)^2} + \dots$ Let's examine the numerators: 8, 12, 16, 20, 24, ... This is an arithmetic progression with the first term $a = 8$ and a common difference $d = 12 - 8 = 4$. The $n^{th}$ term of this arithmetic progression is $a_n = a + (n-1)d = 8 + (n-1)4 = 8 + 4n - 4 = 4n + 4$. Therefore, the $n^{th}$ term of the given series is $\left(\frac{4n+4}{5}\right)^2$.

2. Calculate the Sum of the First Ten Terms ($S_{10}$): We need to find the sum of the first ten terms, so $n$ ranges from 1 to 10. $S_{10} = \sum_{n=1}^{10} \left(\frac{4n+4}{5}\right)^2$ $S_{10} = \sum_{n=1}^{10} \frac{(4(n+1))^2}{5^2}$ $S_{10} = \sum_{n=1}^{10} \frac{16(n+1)^2}{25}$ $S_{10} = \frac{16}{25} \sum_{n=1}^{10} (n+1)^2$

3. Evaluate the Summation of Squares: Let $k = n+1$. When $n=1$, $k=2$. When $n=10$, $k=11$. So, the summation becomes: $\sum_{n=1}^{10} (n+1)^2 = \sum_{k=2}^{11} k^2 = 2^2 + 3^2 + 4^2 + \dots + 11^2$ We can express this sum using the formula for the sum of the first $n$ squares: $\sum_{k=2}^{11} k^2 = \left(\sum_{k=1}^{11} k^2\right) - 1^2$ Using the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$ with $n=11$: $\sum_{k=1}^{11} k^2 = \frac{11(11+1)(2 \times 11 + 1)}{6} = \frac{11(12)(23)}{6}$ $\sum_{k=1}^{11} k^2 = 11 \times 2 \times 23 = 22 \times 23 = 506$ Therefore, $\sum_{k=2}^{11} k^2 = 506 - 1^2 = 506 - 1 = 505$

4. Substitute Back and Find $S_{10}$: Now substitute the value of the summation back into the expression for $S_{10}$: $S_{10} = \frac{16}{25} \times 505$ $S_{10} = \frac{16}{25} \times (5 \times 101)$ $S_{10} = \frac{16 \times 5 \times 101}{25}$ $S_{10} = \frac{16 \times 101}{5}$ $S_{10} = \frac{16}{5} \times 101$

5. Determine the Value of $m$: We are given that the sum of the first ten terms is $\frac{16}{5}m$. We have calculated the sum to be $S_{10} = \frac{16}{5} \times 101$. Equating the two expressions: $\frac{16}{5}m = \frac{16}{5} \times 101$ Dividing both sides by $\frac{16}{5}$, we get: $m = 101$

Common Mistakes & Tips

• Incorrectly Identifying the $n^{th}$ Term: Carefully check the pattern of the numerators and ensure the formula for the $n^{th}$ term is accurate.
• Errors in Sum of Squares Formula Application: Remember that the formula $\sum_{k=1}^{n} k^2$ starts from $k=1$. If the series starts from a different number, adjust the sum accordingly by subtracting the missing terms.
• Arithmetic Errors: Double-check all calculations, especially when dealing with fractions and large numbers.

Summary

The problem involves finding the sum of a series whose terms are squares of numbers forming an arithmetic progression. By converting the mixed fractions to improper fractions, we identified the general term of the series. We then used the formula for the sum of squares of natural numbers to calculate the sum of the first ten terms. Finally, by equating this sum to the given expression $\frac{16}{5}m$, we solved for $m$.

The final answer is \boxed{101}.