Chapter 12.4, Problem 31AYU

Use mathematical induction to prove that

$a\text{ + }\left(a\text{ + }d\right)\text{ + }\left(a\text{ + 2}d\right)\text{ + }\cdot \cdot \cdot \text{ + }\left[a\text{ + }\left(n-1\right)d\right]\text{ = }na\text{ + }d\frac{n\left(n-1\right)}{2}$

To determine

To prove: The given statement $a + \left(a + d\right) + \left(a + 2d\right) + \cdots + \left[a + \left(n - 1\right)d\right] = na + d\frac{n\left(n - 1\right)}{2}$ is true for all natural numbers $n$ using the Principle of Mathematical Induction.

Answer to Problem 31AYU

The statement is true for the natural number $\left($ k + 1$\right)$ terms, it is true for all natural numbers by the theorem of mathematical induction.

Explanation of Solution

Given:

Statements says the $a + \left(a + d\right) + \left(a + 2d\right) + \cdots + \left[a + \left(n - 1\right)d\right] = na + d\frac{n\left(n - 1\right)}{2}$ is true for all natural number.

Formula used:

The Principle of Mathematical Induction

Suppose that the following two conditions are satisfied with regard to a statement about natural numbers:

CONDITION I: The statement is true for the natural number 1.

CONDITION II: If the statement is true for some natural number $k$, it is also true for the next natural number $k + 1$. Then the statement is true for all natural numbers.

Proof:

Consider the statement

$a + \left(a + d\right) + \left(a + 2d\right) + \cdots + \left[a + \left(n - 1\right)d\right] = na + d\frac{n\left(n - 1\right)}{2}$    -----(1)

Step 1: Show that statement (1) is true for the natural number $n = 1$.

That is $a + \left(1 - 1\right)d = 1.a + d\frac{1\left(1 - 1\right)}{2} = a$. Hence the statement is true for the natural number $n = 1$.

Step 2: Assume that the statement is true for some natural number $k$.

That is $a + \left(a + d\right) + \left(a + 2d\right) + \cdots + \left[a + \left(k - 1\right)d\right] = ka + d\frac{k\left(k - 1\right)}{2}$    -----(2)

Step 3: Prove that the statement is true for the next natural number $k + 1$.

That is, to prove that

$a + \left(a + d\right) + \left(a + 2d\right) + \cdots + \left[a + \left(\left(k + 1\right)-1\right)d\right] = \left(k + 1\right)a + d\frac{\left(k + 1\right)\left(\left(k + 1\right) - 1\right)}{2}$

That is $a + \left(a + d\right) + \left(a + 2d\right) + \cdots + \left(a + kd\right) = \left(k + 1\right)a + d\frac{k\left(k + 1\right)}{2}$

Consider $a + \left(a + d\right) + \left(a + 2d\right) + \cdots + \left(a + kd\right)$

There are $\left($ k + 1$\right)$ times of $a$’s are available in the above term.

$a + \left(a + d\right) + \left(a + 2d\right) + \cdots + \left(a + kd\right) = \left(k + 1\right)a + d + 2d + \cdots + kd$

$= \left(k + 1\right)a + \left(1 + 2 + 3 + \cdots + k\right)d$

$= \left(k + 1\right)a + \frac{k\left(k + 1\right)d}{2}$.

As the statement is true for the natural number $\left($ k + 1$\right)$ terms, it is true for all natural numbers by the theorem of mathematical induction.