Question

The positive even natural numbers can be represented as $2 k$, for $k \in \mathrm{Nat}^{+}$. Let $$ \Sigma(k)=2+4+\cdots+2 k $$ be the sum of the first $k$ positive even natural numbers. 1. Write a recursive definition for $\Sigma$ beginning with the base case of $k=1$. 2. Using this recursive definition, prove that $\Sigma(n)=n(n+1)$, for every $n \in \mathrm{Nat}^{+}$.

   The positive even natural numbers can be represented as $2 k$, for $k \in \mathrm{Nat}^{+}$. Let
$$
\Sigma(k)=2+4+\cdots+2 k
$$
be the sum of the first $k$ positive even natural numbers.
1. Write a recursive definition for $\Sigma$ beginning with the base case of $k=1$.
2. Using this recursive definition, prove that $\Sigma(n)=n(n+1)$, for every $n \in \mathrm{Nat}^{+}$.

Logic, sets, and recursion

Robert L. Causey 1st Edition

Chapter 3, Problem 17

Instant Answer

Step 1

- The base case is when $k = 1$. In this case, the sum of the first $k$ positive even natural numbers is simply $2$. Therefore, $\Sigma(1) = 2$. Show more…

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The positive even natural numbers can be represented as $2 k$, for $k \in \mathrm{Nat}^{+}$. Let $$ \Sigma(k)=2+4+\cdots+2 k $$ be the sum of the first $k$ positive even natural numbers. 1. Write a recursive definition for $\Sigma$ beginning with the base case of $k=1$. 2. Using this recursive definition, prove that $\Sigma(n)=n(n+1)$, for every $n \in \mathrm{Nat}^{+}$.

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Key Concepts

Recursive Definition

A recursive definition specifies the value of a function or sequence for any argument by defining it in terms of previous values. In the context of sequences, it starts with one or more base cases and then provides a formula that relates the nth term to one or more preceding terms, thereby allowing the construction of the sequence one step at a time.

Mathematical Induction

Mathematical induction is a proof technique used to establish the truth of an infinite number of cases, typically statements indexed by the natural numbers. It involves proving a base case, and then showing that if the statement holds for an arbitrary case 'n', it must also hold for n+1, thereby demonstrating the truth of the statement for all natural numbers.

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