Key Concepts and Formulas

1. Arithmetic Progression (AP): A sequence where the difference between consecutive terms is constant. The $r$-th term is given by $a_r = a + (r-1)d$, where $a$ is the first term and $d$ is the common difference.

2. Sum of Cubes of First $n$ Natural Numbers: The sum of the cubes of the first $n$ natural numbers is given by the formula: $\sum_{r=1}^{n} r^3 = \left( \frac{n(n+1)}{2} \right)^2$

Step-by-Step Solution

Step 1: Analyze the Given Series and Identify the Pattern in the Bases

The given series is ${\left( {{3 \over 4}} \right)^3} + {\left( {1{1 \over 2}} \right)^3} + {\left( {2{1 \over 4}} \right)^3} + {3^3} + {\left( {3{3 \over 4}} \right)^3} + \ldots$. To understand the structure of the series, we first examine the bases of the cubed terms. We convert all bases to improper fractions with a common denominator of 4: The bases are:

• $\frac{3}{4}$
• $1\frac{1}{2} = \frac{3}{2} = \frac{6}{4}$
• $2\frac{1}{4} = \frac{9}{4}$
• $3 = \frac{12}{4}$
• $3\frac{3}{4} = \frac{15}{4}$

The sequence of bases is $\frac{3}{4}, \frac{6}{4}, \frac{9}{4}, \frac{12}{4}, \frac{15}{4}, \ldots$. This sequence is an Arithmetic Progression (AP) with the first term $a = \frac{3}{4}$ and a common difference $d = \frac{6}{4} - \frac{3}{4} = \frac{3}{4}$.

Step 2: Determine the General Term ($r$-th term) of the Series

The $r$-th term of the AP of bases is given by $a_r = a + (r-1)d$. Substituting $a = \frac{3}{4}$ and $d = \frac{3}{4}$: $a_r = \frac{3}{4} + (r-1)\frac{3}{4} = \frac{3}{4}(1 + r - 1) = \frac{3r}{4}$ The $r$-th term of the given series, $T_r$, is the cube of this base: $T_r = \left( \frac{3r}{4} \right)^3$

Step 3: Formulate the Sum of the First 15 Terms

We need to find the sum of the first 15 terms, $S_{15}$. This is given by: $S_{15} = \sum_{r=1}^{15} T_r = \sum_{r=1}^{15} \left( \frac{3r}{4} \right)^3$ We can simplify the expression inside the summation: $S_{15} = \sum_{r=1}^{15} \frac{3^3 r^3}{4^3} = \sum_{r=1}^{15} \frac{27 r^3}{64}$ We can factor out the constant $\frac{27}{64}$: $S_{15} = \frac{27}{64} \sum_{r=1}^{15} r^3$

Step 4: Apply the Formula for the Sum of Cubes

Using the formula for the sum of the first $n$ cubes, $\sum_{r=1}^{n} r^3 = \left( \frac{n(n+1)}{2} \right)^2$, with $n=15$: $\sum_{r=1}^{15} r^3 = \left( \frac{15(15+1)}{2} \right)^2 = \left( \frac{15 \times 16}{2} \right)^2$ $\sum_{r=1}^{15} r^3 = (15 \times 8)^2 = (120)^2 = 14400$

Step 5: Calculate the Total Sum and Solve for $k$

Substitute the sum of cubes back into the expression for $S_{15}$: $S_{15} = \frac{27}{64} \times 14400$ To simplify, we divide 14400 by 64: $\frac{14400}{64} = \frac{144 \times 100}{64} = \frac{9 \times 16 \times 100}{4 \times 16} = \frac{9 \times 100}{4} = 9 \times 25 = 225$ So, $S_{15} = 27 \times 225$ The problem states that $S_{15} = 225k$. Equating our result: $27 \times 225 = 225k$ Dividing both sides by 225 to solve for $k$: $k = \frac{27 \times 225}{225} = 27$

Common Mistakes & Tips

• Inconsistent Base Representation: Ensure all bases are converted to a uniform format (like improper fractions with a common denominator) to easily spot the AP.
• Algebraic Errors: Be meticulous with algebraic manipulations, especially when factoring and simplifying fractions within summations.
• Formula Recall: Have the standard summation formulas for $r$, $r^2$, and $r^3$ readily available.

Summary

The problem involves finding the sum of a series where each term is the cube of terms in an arithmetic progression. By identifying the arithmetic progression of the bases, we derived the general term of the series. We then used the formula for the sum of cubes of natural numbers to calculate the sum of the first 15 terms. Finally, by equating this sum to the given expression $225k$, we solved for $k$.

The final answer is $\boxed{27}$.