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Magazine Chapter 3: Problem 3
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
Short Answer
Expert verified
Options b and d are correct.
Step by step solution
01
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 \( 2r^3 + 5r^2 + 3r \).
02
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:** \[ \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \].
03
Find Each Sum Component
We substitute and simplify each component:\( \sum_{r=1}^{n} 2r^3 = 2 \cdot \left(\frac{n(n+1)}{2}\right)^2 = \frac{n^2(n+1)^2}{2} \)\( \sum_{r=1}^{n} 5r^2 = 5 \cdot \frac{n(n+1)(2n+1)}{6} = \frac{5n(n+1)(2n+1)}{6} \)\( \sum_{r=1}^{n} 3r = 3 \cdot \frac{n(n+1)}{2} = \frac{3n(n+1)}{2} \).
04
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:\[ \frac{3n^2(n+1)^2}{6} + \frac{5n(n+1)(2n+1)}{6} + \frac{9n(n+1)}{6}. \]
05
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 \(a = \frac{1}{2}, b = \frac{8}{3}, c = \frac{9}{2}, d = \frac{7}{3}, e = 0 \).
06
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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.
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.
