Key Concepts and Formulas

• Sum of Cubes: The sum of the cubes of the first $n$ natural numbers is given by the formula: $\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2$
• Sum of Squares: The sum of the squares of the first $n$ natural numbers is given by the formula: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$
• Sum of Natural Numbers: The sum of the first $n$ natural numbers is given by the formula: $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$
• General Term of an Arithmetic Progression: The $k$-th term of an arithmetic progression is $a + (k-1)d$, where $a$ is the first term and $d$ is the common difference.

Step-by-Step Solution

We are given the equation: $\frac{1^3 + 2^3 + 3^3 + \dots \text{ up to } n \text{ terms}}{1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \dots \text{ up to } n \text{ terms}} = \frac{9}{5}$

Step 1: Evaluate the Sum of the Numerator The numerator is the sum of the cubes of the first $n$ natural numbers. Using the formula for the sum of cubes: $\text{Numerator} = \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2$

Step 2: Determine the General Term of the Denominator Series Let's examine the terms in the denominator: $1 \cdot 3, 2 \cdot 5, 3 \cdot 7, \dots$ For the $k$-th term: The first factor is $k$. The second factor follows the sequence $3, 5, 7, \dots$. This is an arithmetic progression with the first term $a=3$ and a common difference $d=2$. The $k$-th term of this AP is $3 + (k-1)2 = 3 + 2k - 2 = 2k+1$. Therefore, the general $k$-th term of the denominator series is $T_k = k(2k+1) = 2k^2 + k$.

Step 3: Evaluate the Sum of the Denominator The sum of the denominator is the sum of the general terms from $k=1$ to $n$: $\text{Denominator} = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (2k^2 + k)$ Using the linearity of summation and the standard formulas: $\text{Denominator} = 2 \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k$ $\text{Denominator} = 2 \left(\frac{n(n+1)(2n+1)}{6}\right) + \frac{n(n+1)}{2}$ Simplify the expression: $\text{Denominator} = \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2}$ Factor out the common term $n(n+1)$: $\text{Denominator} = n(n+1) \left(\frac{2n+1}{3} + \frac{1}{2}\right)$ Find a common denominator for the terms in the parenthesis: $\text{Denominator} = n(n+1) \left(\frac{2(2n+1) + 3(1)}{6}\right)$ $\text{Denominator} = n(n+1) \left(\frac{4n+2+3}{6}\right)$ $\text{Denominator} = \frac{n(n+1)(4n+5)}{6}$

Step 4: Substitute the Sums into the Given Equation and Solve for $n$ Now, substitute the expressions for the numerator and denominator into the given equation: $\frac{\left(\frac{n(n+1)}{2}\right)^2}{\frac{n(n+1)(4n+5)}{6}} = \frac{9}{5}$ Simplify the numerator: $\frac{\frac{n^2(n+1)^2}{4}}{\frac{n(n+1)(4n+5)}{6}} = \frac{9}{5}$ Cancel out common factors of $n(n+1)$ (since $n \ge 1$, $n \neq 0$ and $n+1 \neq 0$): $\frac{\frac{n(n+1)}{4}}{\frac{4n+5}{6}} = \frac{9}{5}$ Multiply the numerator by the reciprocal of the denominator: $\frac{n(n+1)}{4} \cdot \frac{6}{4n+5} = \frac{9}{5}$ Simplify the constants: $\frac{3n(n+1)}{2(4n+5)} = \frac{9}{5}$ Cross-multiply: $5 \cdot 3n(n+1) = 9 \cdot 2(4n+5)$ $15n(n+1) = 18(4n+5)$ Expand both sides: $15n^2 + 15n = 72n + 90$ Rearrange into a quadratic equation: $15n^2 + 15n - 72n - 90 = 0$ $15n^2 - 57n - 90 = 0$ Divide by the common factor of 3: $5n^2 - 19n - 30 = 0$ Factor the quadratic equation. We look for two numbers that multiply to $5 \times -30 = -150$ and add to $-19$. These numbers are $-25$ and $6$. $5n^2 - 25n + 6n - 30 = 0$ Factor by grouping: $5n(n - 5) + 6(n - 5) = 0$ $(5n + 6)(n - 5) = 0$ This yields two possible solutions for $n$: $5n + 6 = 0 \implies n = -\frac{6}{5}$ $n - 5 = 0 \implies n = 5$

Step 5: Select the Valid Solution Since $n$ represents the number of terms in a series, it must be a positive integer. Therefore, $n = -\frac{6}{5}$ is an extraneous solution. The valid solution is $n = 5$.

Common Mistakes & Tips

• Incorrect General Term: Carefully derive the general term for the denominator series by identifying the pattern in both factors of each term.
• Algebraic Errors: Be meticulous with algebraic manipulations, especially when simplifying fractions and solving the quadratic equation.
• Extraneous Roots: Always check if the obtained values of $n$ are valid in the context of the problem (e.g., $n$ must be a positive integer for a number of terms).

Summary The problem requires the evaluation of two series sums. The numerator is the sum of cubes, for which a standard formula exists. The denominator's general term was identified as $k(2k+1)$, and its sum was calculated using formulas for the sum of squares and sum of natural numbers. Equating the ratio of these sums to $\frac{9}{5}$ led to a quadratic equation in $n$. Solving this equation and selecting the positive integer solution gives the final answer.

The final answer is \boxed{5}.