Problem 70 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

For any odd integer \(n \geq 1\), \(n^{3}-(n-1)^{3}+\ldots+(-1)^{n-1} 1^{3}=\) (A) \(\frac{1}{2}(n-1)^{2}(2 n-1)\) (B) \(\frac{1}{4}(n-1)^{2}(2 n-1)\) (C) \(\frac{1}{2}(n+1)^{2}(2 n-1)\) (D) \(\frac{1}{4}(n+1)^{2}(2 n-1)\)

Step-by-Step Solution

Answer

The solution is (D) \(\frac{1}{4}(n+1)^2(2n-1)\).

Step 1: Recognize the pattern and rewrite the sum

Recognize that the expression is an alternating sum of cubes of consecutive odd integers starting from \(n^3\) down to \((-1)^{n-1} 1^3\). The alternating signs suggest a telescoping series. The given series is \[n^3 - (n-1)^3 + (n-2)^3 - (n-3)^3 + \ldots + (-1)^{n-1}\times 1^3.\]

Step 2: Simplify each group in the sequence

Rewrite the sequence by grouping terms into pairs: \[(n^3 - (n-1)^3) + ((n-2)^3 - (n-3)^3) + \ldots\] where each pair is of the form \((k^3 - (k-1)^3)\).

Step 3: Compute each pair (k^3 - (k-1)^3)

Compute the difference of cubes for each pair: \[ k^3 - (k-1)^3 = k^3 - (k^3 - 3k^2 + 3k - 1) \]This simplifies to \[ 3k^2 - 3k + 1.\] This represents the contribution from each pair in the sequence.

Step 4: Sum the expanded pair contributions

For odd \(n\), the number of terms is \((n+1)/2\). The sum of squares is given by \(\sum_{k=1}^{m} k^2 = \frac{m(m+1)(2m+1)}{6}\). Substitute \(m = (n+1)/2\) into the formula:\[\sum_{k=1}^{\frac{n+1}{2}} 3k^2 - 3k + 1 = 3 \sum_{k=1}^{\frac{n+1}{2}} k^2 - 3 \sum_{k=1}^{\frac{n+1}{2}} k + \frac{n+1}{2}\].

Step 5: Calculate each component

First calculate the sum of squares:\[ \sum_{k=1}^{m} k^2 = \frac{m(m+1)(2m+1)}{6}.\]Substitute \(m = \frac{n+1}{2}\) and evaluate it similarly for the linear sum \( \sum_{k=1}^{m} k = \frac{m(m+1)}{2}\).

Step 6: Find terms from simplified contributions

Combine and simplify results from prior steps:For \(m = \frac{n+1}{2}\):- Sum of squares becomes \(\frac{(n+1)(n+2)(2n+3)}{24}\)- Sum of linear becomes \(\frac{(n+1)(n+2)}{4}\)- Add \(\frac{n+1}{2} \) from the unit term count.These results combine for the polynomial expression.

Step 7: Simplify final expression

Through algebraic simplification using these computed sums, simplify to find:The series simplifies collectively to \(\frac{1}{4}(n+1)^2(2n-1)\).

Key Concepts

Difference of Cubes

When we talk about the difference of cubes, we're referring to the mathematical expression of the form \( a^3 - b^3 \). This is a recognizable pattern, often used in algebra for simplification and factorization. The difference of cubes can be factored using the formula:

• \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \)

This pattern arises in various calculations including those involving polynomials where terms are expressed as cubes.
In the context of this exercise, the series \( n^3 - (n-1)^3 \) represents a sequence of such differences, highlighting the use of the difference of cubes within a telescoping series, acting as a simplified step in solving the problem.

Sum of Squares

The sum of squares deals with the sum of square terms like \( k^2 \), where \( k \) runs over a range of integers. The formula for the sum of squares of the first \( m \) natural numbers is:

• \( \sum_{k=1}^{m} k^2 = \frac{m(m+1)(2m+1)}{6} \)

This formula is useful in breaking down complex series into manageable units.
In solving the original exercise, the sum of squares formula is crucial for the computation of \( \sum_{k=1}^{\frac{n+1}{2}} 3k^2 \), which is part of simplifying the contributions from the alternating terms of the series. Identifying these patterns helps in finding a structured and easier path to solve seemingly complex algebraic series.

Telescoping Series

A telescoping series is a series where consecutive terms cancel each other out partially or completely. This phenomenon makes it much easier to find the sum because once expanded, much of the series collapses to fewer terms.
The telescoping nature of the original series, \( n^3 - (n-1)^3 + \dots + (-1)^{n-1}1^3 \), simplifies the solution significantly by letting terms cancel sequentially, leaving a limited number of terms to compute explicitly. In practice, when solving such problems, identifying the potential for telescoping can be immensely beneficial. This not only reduces complex calculations but also avoids potential errors associated with adding long series.