(2³ - 1³)/(1 × 7) + (4³ - 3³ + 2³ - 1³)/(2 × 11) + (6³ - 5³ + 4³ - 3³ + 2³ - 1³)/(3 × 15) + ... + (3

Q: Wait, why do we reject the negative root here? Or why can't we just use energy conservation?

In this specific series problem, we are dealing with discrete summation, not a physical system where energy conservation applies. We are simply simplifying a complex algebraic expression. There are no roots to reject because we aren't solving a quadratic equation for a variable; we are evaluating a sum.

Q: Why did the $8k^3$ cancel out? Did I make a mistake?

You did not make a mistake! The cancellation of the $8k^3$ terms is the 'magic' of this problem. When you expand $(2k-1)^3$ as $8k^3 - 12k^2 + 6k - 1$ and subtract it from $(2k)^3$, the $8k^3$ terms subtract to zero, leaving behind a much simpler quadratic. It is designed to collapse!

Q: How do I know the number of terms is 15?

Look at the denominator of the last term: $15 \times 63$. We established that the general denominator is $n(4n+3)$. If we set $n=15$, we get $15 \times (4(15)+3) = 15 \times 63$. This confirms that the series ends exactly at the 15th term.

Question

$\frac{2^3 - 1^3}{1 	imes 7} + \frac{4^3 - 3^3 + 2^3 - 1^3}{2 	imes 11} + \frac{6^3 - 5^3 + 4^3 - 3^3 + 2^3 - 1^3}{3 	imes 15} + \dots + \frac{30^3 - 29^3 + 28^3 - 27^3 + \dots + 2^3 - 1^3}{15 	imes 63}$ is equal to ________.

Sigma Percentile · Accepted Answer

### The Beauty of Hidden Patterns

Imagine you are standing before a massive, intimidating wall of numbers. You see cubes, fractions, and a series that seems to stretch into infinity.

It is easy to feel overwhelmed. But in the world of JEE Advanced, these problems are not designed to break you; they are designed to test your ability to see the order within the chaos. Let us embark on a journey to dismantle this series, step by step.

### Phase 1: Decoding the Denominator

First, look at the denominators: $1 	imes 7$, $2 	imes 11$, $3 	imes 15$, and so on. If you stare at them long enough, a rhythm emerges.

The first factor is simply the term number, $n$. The second factor is $4n+3$.

So, the denominator of our general term, $T_n$, is $n(4n+3)$. By identifying this, we have already taken the first step toward victory. We have turned a chaotic list into a predictable mathematical rule.

### Phase 2: The Numerator's Secret

Now, let us tackle the numerators. They are built from differences of cubes. For the $n$-th term, the numerator $N_n$ is the sum of $(2k)^3 - (2k-1)^3$ for $k$ from $1$ to $n$.

This looks terrifying, but let us apply the algebraic identity $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$. Expanding $(2k-1)^3$ gives us $8k^3 - 12k^2 + 6k - 1$.

When we subtract this from $(2k)^3$, the $8k^3$ terms vanish into thin air! We are left with the elegant quadratic: $12k^2 - 6k + 1$.

### Phase 3: The Summation Magic

With the numerator simplified to $12k^2 - 6k + 1$, we can now use our standard summation tools. We distribute the sum:

$$ \sum_{k=1}^{n} (12k^2 - 6k + 1) = 12 \sum_{k=1}^{n} k^2 - 6 \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 $$

Using the standard formulas:

$$ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \quad 	ext{and} \quad \sum_{k=1}^{n} k = \frac{n(n+1)}{2} $$

After some careful algebraic factoring, we find that the numerator simplifies to:

$$ N_n = n^2(4n+3) $$

### Phase 4: The Grand Cancellation

This is the moment of truth. We have the general term defined as:

$$ T_n = \frac{N_n}{	ext{Denominator}} = \frac{n^2(4n+3)}{n(4n+3)} $$

The $(4n+3)$ terms cancel out, and one $n$ cancels out. We are left with the simple result:

$$ T_n = n $$

All that complexity, all those cubes, reduced to the simple, humble integer $n$. The series is just the sum of the first 15 natural numbers.

The final calculation is:

$$ \sum_{n=1}^{15} n = \frac{15 	imes 16}{2} = 120 $$

This is the reward for our persistence. Remember, in mathematics, complexity is often just a mask for a deeper, simpler truth. The final answer is **120**.

Enter Numerical Value:

Visualized Solution

Observe the Series Pattern

Define the Numerator of $T_{n}$

Expand the Cubic Term $(2 k - 1)^{3}$

Simplify the Summand

Apply Summation Formulas

Substitute into Numerator

Factor Out Common Terms

Expand Inside the Bracket

Final Form of Numerator

Simplify the General Term $T_{n}$

Identify the Number of Terms

Calculate the Total Sum

The Sigma Insight: Sum of Special Series

The Beauty of Hidden Patterns

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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