If (1³ + 2³ + 3³ + ...upto n terms) / (1·3 + 2·5 + 3·7 + ...upto n terms) = 9/5, then the value of n

Q: Why did we write the general term of the denominator as $r(2r+1)$?

Great observation! If you look at the sequence $1 \cdot 3, 2 \cdot 5, 3 \cdot 7, \dots$, the first factor is simply $r$. The second factor follows the arithmetic progression $3, 5, 7, \dots$, which has a first term $a=3$ and common difference $d=2$. Thus, the $r$-th term of the second factor is $3 + (r-1)2 = 2r+1$. Multiplying them gives $r(2r+1)$.

Q: Can I just expand everything and solve for $n$?

While technically possible, it is a trap! Expanding polynomials of degree 3 or 4 will lead to massive algebraic errors. Always look for common factors like $n(n+1)$ to factor out before simplifying the ratio. It makes the math elegant and much faster.

Q: Why do we reject $n = -\frac{6}{5}$?

In the context of this problem, $n$ represents the number of terms in a sequence. By definition, $n$ must be a positive integer ($n \in \mathbb{N}$). A negative or fractional number of terms is physically and mathematically meaningless here.

Q: Is there a shortcut to sum the denominator?

You could use the method of differences or properties of summation, but the standard approach of splitting $2r^2 + r$ into two separate sums is the most reliable way to avoid mistakes in JEE Advanced. Stick to the standard formulas!

Q: What if I get stuck during the algebraic simplification of the ratio?

Take a deep breath and look for common terms in the numerator and denominator. Notice that $\frac{n(n+1)}{2}$ appears in both. Dividing by these common factors early on is the secret to turning a complex-looking equation into a simple quadratic.

Question

If $\frac{1^3 + 2^3 + 3^3 + \dots 	ext{upto } n 	ext{ terms}}{1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \dots 	ext{upto } n 	ext{ terms}} = \frac{9}{5}$, then the value of $n$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE excellence! Today, we are not just solving an algebraic equation; we are peeling back the layers of a mathematical structure.

Imagine you are standing before a grand staircase, where each step is built upon the logic of sequences and series. Our goal is to find the specific step $n$ that balances the ratio of two complex sums to exactly $\frac{9}{5}$.

### Decoding the Numerator

Let us look at the numerator: $1^3 + 2^3 + 3^3 + \dots + n^3$. This is a classic identity that every aspirant should hold dear.

We know that the sum of the first $n$ cubes is given by the square of the sum of the first $n$ natural numbers. Mathematically, we express this as:

$$	ext{Numerator} = \sum_{r=1}^n r^3 = \left[ \frac{n(n+1)}{2} ight]^2 = \frac{n^2(n+1)^2}{4}$$

This is our foundation. It is clean, elegant, and ready to be used.

### The Anatomy of the Denominator

Now, let us turn our attention to the denominator: $1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \dots$. At first glance, it looks intimidating, but the pattern is clear.

The first factor is simply $r$, and the second factor is an arithmetic progression: $3, 5, 7, \dots$. The $r$-th term of this progression is $2r+1$. Thus, the general term $T_r$ is:

$$T_r = r(2r+1) = 2r^2 + r$$

To find the total sum, we apply the summation operator across this expression:

$$	ext{Denominator} = \sum_{r=1}^n (2r^2 + r) = 2\sum_{r=1}^n r^2 + \sum_{r=1}^n r$$

### The Power of Standard Formulas

Here is where we pull from our toolkit. We substitute the standard formulas for $\sum r^2$ and $\sum r$:

$$	ext{Denominator} = 2 \left[ \frac{n(n+1)(2n+1)}{6} ight] + \frac{n(n+1)}{2}$$

By simplifying this, we obtain:

$$	ext{Denominator} = \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2}$$

To make this manageable, we factor out the common term $\frac{n(n+1)}{2}$:

$$	ext{Denominator} = \frac{n(n+1)}{2} \left[ \frac{2(2n+1)}{3} + 1 ight] = \frac{n(n+1)(4n+5)}{6}$$

### The Grand Convergence

Now, we bring the numerator and denominator together. We are given that the ratio is $\frac{9}{5}$:

$$\frac{\frac{n^2(n+1)^2}{4}}{\frac{n(n+1)(4n+5)}{6}} = \frac{9}{5}$$

Watch closely as the terms $n(n+1)$ cancel out, leaving us with:

$$\frac{3n(n+1)}{2(4n+5)} = \frac{9}{5} \Rightarrow \frac{n(n+1)}{2(4n+5)} = \frac{3}{5}$$

This transforms into a simple quadratic equation:

$$5(n^2 + n) = 6(4n + 5) \Rightarrow 5n^2 - 19n - 30 = 0$$

Solving this quadratic, we find $(n-5)(5n+6) = 0$. Since $n$ must be a natural number, we discard the negative root.

Thus, **$n=5$**. You have successfully navigated the complexity and arrived at the truth.

Animated Solution for Mathematics - Sequence and Series: If $\frac{1 ^{3} + 2 ^{3} + 3 ^{3} + \dots upto n terms}{1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \dots upto n terms} = \frac{9}{5}$ , then the value of $n$ is

Enter Numerical Value:

Visualized Solution

Analyzing the Numerator: Sum of Cubes

Identifying the Denominator's General Term

Setting up the Denominator Summation

Substituting Standard Summation Formulas

Factoring the Denominator Expression

Final Simplified Denominator

Setting up the Ratio Equation

Algebraic Cancellation and Reduction

Forming the Quadratic Equation

Solving the Quadratic for n

The Final Answer

The Sigma Insight: Sum of Special Series

Analyzing the Setup

Decoding the Numerator

The Anatomy of the Denominator

The Power of Standard Formulas

The Grand Convergence

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