91Ó°ÊÓ

The objective of this problem is to derive a formula for \(\sum_{k=1}^{n} k^{2}\). (a) Use induction to show that $$ \sum_{k=1}^{n}\left(\begin{array}{l} k \\ 1 \end{array}\right)=\frac{n(n+1)}{2} $$ for any positive integer \(n\). (b) Use induction to show that $$ \sum_{k=1}^{n}\left(\begin{array}{l} k \\ 2 \end{array}\right)=\frac{n(n+1)(n-1)}{3 !} $$ for any positive integer \(n\). Hint: Note that \(\left(\begin{array}{l}1 \\\ 2\end{array}\right)=0\) (c) Find the integers \(a\) and \(b\) such that $$ k^{2}=a\left(\begin{array}{l} k \\ 2 \end{array}\right)+b\left(\begin{array}{l} k \\ 1 \end{array}\right) $$ Hint: Compare coefficients. (d) From part (c), we obtain $$ \sum_{k=1}^{n} k^{2}=a \sum_{k=1}^{n}\left(\begin{array}{l} k \\ 2 \end{array}\right)+b \sum_{k=1}^{n}\left(\begin{array}{l} k \\ 1 \end{array}\right) $$ Apply the results from parts (a) and (b) to derive a formula for \(\sum_{k=1}^{n} k^{2}\).

Step by step solution

01

Prove Formula for Sum of First k Using Induction

We want to prove \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \) using induction.1. **Base case**: For \( n = 1 \), \( \sum_{k=1}^{1} k = 1 \). The formula gives \( \frac{1(1+1)}{2} = 1 \), so the base case holds.2. **Inductive step**: Assume true for \( n = m \), i.e., \( \sum_{k=1}^{m} k = \frac{m(m+1)}{2} \). For \( n = m+1 \),\[\sum_{k=1}^{m+1} k = \sum_{k=1}^{m} k + (m+1) = \frac{m(m+1)}{2} + (m+1).\]Simplifying gives:\[\frac{m(m+1)}{2} + (m+1) = \frac{m(m+1) + 2(m+1)}{2} = \frac{(m+1)(m+2)}{2} = \frac{(m+1)((m+1)+1)}{2}.\]Thus, the inductive step holds, proving the formula by induction.

02

Prove Formula for Sum of Combinations Using Induction

We need to prove \( \sum_{k=1}^{n} \binom{k}{2} = \frac{n(n+1)(n-1)}{6} \) using induction.1. **Base case**: For \( n = 1 \), \( \sum_{k=1}^{1} \binom{k}{2} = 0 \) since \( \binom{1}{2} = 0 \). The formula gives \( \frac{1(1+1)(1-1)}{6} = 0 \), so the base case holds.2. **Inductive step**: Assume true for \( n = m \), i.e., \( \sum_{k=1}^{m} \binom{k}{2} = \frac{m(m+1)(m-1)}{6} \). For \( n = m+1 \),\[\sum_{k=1}^{m+1} \binom{k}{2} = \sum_{k=1}^{m} \binom{k}{2} + \binom{m+1}{2} = \frac{m(m+1)(m-1)}{6} + \frac{m(m+1)}{2}.\]Simplifying gives:\[\frac{m(m+1)(m-1) + 3m(m+1)}{6} = \frac{m(m+1)(m+2)}{6},\]matching the formula for \( n = m+1 \). Thus, the inductive step holds, completing the proof.

03

Find Coefficients for Expression of k^2

We want to express \( k^2 \) as a linear combination of binomial coefficients:\[k^2 = a\binom{k}{2} + b\binom{k}{1}.\]Expanding the binomials, we have:\[a\left(\frac{k(k-1)}{2}\right) + b\cdot k = \frac{a}{2}(k^2 - k) + bk.\]Equating coefficients with \( k^2 = \frac{a}{2}k^2 - \frac{a}{2}k + bk \):- Coefficient of \( k^2 \): \( \frac{a}{2} = 1 \), so \( a = 2 \).- Coefficient of \( k \): \( -\frac{a}{2} + b = 0 \), so \( -1 + b = 0 \) or \( b = 1 \).Thus, \( a = 2 \) and \( b = 1 \).

04

Derive the Formula for Sum of k^2

Using the expression from Step 3: \[\sum_{k=1}^{n} k^2 = 2\sum_{k=1}^{n}\binom{k}{2} + \sum_{k=1}^{n}\binom{k}{1}.\]Insert known formulas from Steps 1 and 2:\[2\left(\frac{n(n+1)(n-1)}{6}\right) + \frac{n(n+1)}{2}.\]Simplifying:\[\frac{n(n+1)(n-1)}{3} + \frac{n(n+1)}{2} = \frac{2n(n+1)(n-1) + 3n(n+1)}{6} = \frac{n(n+1)(2n - 1)}{6}.\]Hence, the formula for \( \sum_{k=1}^{n} k^2 \) is \( \frac{n(n+1)(2n+1)}{6} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sum of Squares Formula

The sum of squares formula allows us to quickly calculate the sum of the squares of the first \( n \) natural numbers. It's represented by the equation:\[\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\]This formula is a handy tool in various mathematical fields. It provides an efficient way to handle large sums of squares without performing all multiplication and addition steps one by one.
Understanding and deriving it requires knowledge of binomial coefficients and induction, which are fundamental concepts in algebra.

• It's applicable in statistics, physics, and even computer science.
• Provides quick computation for sums up to a given number.

Binomial Coefficients

Binomial coefficients appear frequently in algebra and combinatorics. They are expressed as \( \binom{n}{k} \) and represent the number of ways to choose \( k \) elements from a set of \( n \) elements without regard to order.
They can be calculated directly using the formula:\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]Where \( ! \) denotes factorial, the product of all positive integers up to that number.

• Used in expansions of expressions raised to a power, such as \((a + b)^n\).
• Integral to probability and statistics, providing formulas for distribution computations.
• Helps in deriving formulas involving combinations and permutations.

Mathematical Induction

Mathematical induction is a method of proof used in mathematics to establish the truth of an infinite number of statements. This technique is akin to falling dominoes: knocking the first one down sets off a chain reaction.

• **Base Case**: Prove the statement for the initial value (usually \( n = 1 \)).
• **Inductive Step**: Assume the statement is true for \( n = k \), and then prove it for \( n = k+1 \).

If both steps are successfully completed, the statement is considered proven for all natural numbers. In the exercise, induction was used to validate parts (a) and (b). This ensures mathematical accuracy across infinite scenarios.

Discrete Mathematics

Discrete mathematics deals with distinct and separated values. Unlike continuous mathematics which works with related variables like calculus, discrete mathematics focuses on integers and finite systems.

• Involves subjects like graph theory, set theory, and combinatorics.
• Highly relevant in computer science, especially in algorithms and data structures.

The application of discrete mathematics in proofs using induction shows its relevance in logical reasoning, cryptography, and network analysis.

Recursive Formulas

Recursive formulas express each term of a sequence as a function of the preceding terms. In mathematics, they help in constructing sequences systematically rather than computing terms individually.

• Useful in computer algorithms, especially those dealing with iterative or repetitive tasks.
• Helps simplify complex calculations by breaking them down into simpler, repeatable steps.

In the context of our exercise, recursive methods complemented by induction help establish relationships between terms of different mathematical problems, thus aiding in deriving formulas like the sum of squares.