Problem 3 - Algebra for JEE ADVANCED

Question

If \(\sum_{r=1}^{n} r(r+1)(2 r+3)=a n^{4}+b n^{3}+c n^{2}+d n+e\), then a. \(a-b=d-c\) b. \(e=0\) c. \(a, b-2 / 3, c-1\) are in A.P. d. \((b+d) / a\) is an integer

Step-by-Step Solution

Answer

Options b and d are correct.

Step 1: Expand the General Term

We begin by expanding the expression for the general term of the summation: \[ r(r+1)(2r+3) = r(2r^2 + 3r + 2r + 3) = 2r^3 + 5r^2 + 3r. \]So, the general term to sum becomes \( 2r^3 + 5r^2 + 3r \).

Step 2: Apply Summation Identity

Next, we apply the formula for the sum of powers:**Sum of cubes:** \[ \sum_{r=1}^{n} r^3 = \left(\frac{n(n+1)}{2}\right)^2 \]**Sum of squares:** \[ \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \]**Sum of first n numbers:** \[ \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \].

Step 3: Find Each Sum Component

We substitute and simplify each component:\( \sum_{r=1}^{n} 2r^3 = 2 \cdot \left(\frac{n(n+1)}{2}\right)^2 = \frac{n^2(n+1)^2}{2} \)\( \sum_{r=1}^{n} 5r^2 = 5 \cdot \frac{n(n+1)(2n+1)}{6} = \frac{5n(n+1)(2n+1)}{6} \)\( \sum_{r=1}^{n} 3r = 3 \cdot \frac{n(n+1)}{2} = \frac{3n(n+1)}{2} \).

Step 4: Compute the Entire Sum

The overall expression for the sum is:\[ \sum_{r=1}^{n} [2r^3 + 5r^2 + 3r] = \frac{n^2(n+1)^2}{2} + \frac{5n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{2}. \]We need to combine these fractions over a common denominator, which is 6. Reformulating gives us:\[ \frac{3n^2(n+1)^2}{6} + \frac{5n(n+1)(2n+1)}{6} + \frac{9n(n+1)}{6}. \]

Step 5: Simplify and Match Coefficients

Simplifying the expression gives:\[ \frac{3n^4 + 6n^3 + 3n^2 + 10n^3 + 15n^2 + 5n + 9n^2 + 9n}{6}. \]Combine like terms: \[ = \frac{3n^4 + 16n^3 + 27n^2 + 14n}{6}. \]Ultimately, compare coefficients to match with \( a n^4 + b n^3 + c n^2 + d n + e \), resulting in \(a = \frac{1}{2}, b = \frac{8}{3}, c = \frac{9}{2}, d = \frac{7}{3}, e = 0 \).

Step 6: Validate Each Option

a. Check if \(a - b = d - c\): \( \frac{1}{2} - \frac{8}{3} eq \frac{7}{3} - \frac{9}{2} \).b. Check if \(e = 0\): Yes.c. Check if \(a, b - \frac{2}{3}, c - 1\) are in A.P: No.d. Check if \((b + d) / a\) is an integer: \((\frac{8}{3} + \frac{7}{3}) / \frac{1}{2} = 10\), which is indeed an integer.

Key Concepts

Sum of Cubes

The sum of cubes relates to the series \( \sum_{r=1}^{n} r^3 \), which represents the total of each integer cubed from 1 to n. This particular summation has a neat formula: \[ \left(\frac{n(n+1)}{2}\right)^2 \]. This looks like an arithmetic progression squared, but arranged cubically.

• This formula stems from the idea that the sum of cubes can be seen as stacking cubes in layers, similar to a 3D pyramid.
• Recognizing this pattern can simplify solving problems that involve cubed terms.
• It helps convert a seemingly complex cube-term series into a familiar polynomial that is more manageable.

For any student tackling problems involving cubes, understanding this concept is essential as it simplifies the calculation by providing a direct formula.

Sum of Squares

The sum of squares summarization pertains to the formula \( \sum_{r=1}^{n} r^2 \), which totalizes every whole number's square from 1 to n. The formula for this sum is: \[ \frac{n(n+1)(2n+1)}{6} \]. This formula sums each squared number consecutively.

• Think of the sum of squares as a method to account for particular arrangements in 2D that layers each successive square on the last one.
• This formula is frequently part of algebraic problems and simplifying complex polynomial operations.

It plays a critical role in algebra because it efficiently sums squared numbers, offering a quick way to reach the result without listing or calculating every square individually.

Arithmetic Progression

Arithmetic progression (A.P.) refers to a sequence of numbers in which each term after the first is derived by adding a fixed number, called the common difference, to the previous term.

• In this context, recognizing sequences as A.P. allows us to efficiently predict future numbers and understand patterns in series.
• The relevance of A.P. in the context of polynomial expression lies in checking if calculated coefficients follow such progressions, revealing underlying patterns or symmetries.
• The formula for the nth term in an arithmetic progression is \( a_n = a_1 + (n-1) \cdot d \), where \( a_1 \) is the first term and \( d \) is the common difference.

Understanding arithmetic progression helps simplify the solving process of polynomial equations by revealing relationships between terms.

Polynomial Coefficients

Polynomial coefficients in an expression like \( a n^4 + b n^3 + c n^2 + d n + e \) represent the multipliers of the variable terms, indicating their influence on the polynomial's behavior.

• These coefficients are crucial because they determine the polynomial's shape when graphed and describe its roots, intercepts, and turning points.
• In the context of algebraic summations, calculating and matching these coefficients allows us to break down complex expressions into understandable terms.
• Comparing coefficients, as done in the exercise, allows for checking algebraic identities and deriving relationships between different parts of the polynomial.

Clear understanding of polynomial coefficients aids in solving, analyzing, and graphically representing polynomial equations, making it easier to interpret results.