Chapter 12, Problem 51SGA

To determine

To prove: that the given proposition is valid for all positive integral values of $n$ by using mathematical induction.

Explanation of Solution

Given information:

  $1+2+3+\mathrm{...}+n=\frac{n\left(n+1\right)}{2}$

Calculation:

Let the given proposition be $S_n$ .

  $S_n=\frac{n\left(n+1\right)}{2}$

Now verify that $S_n$ is valid for $n=1$ .

  $\begin{array}{c}{S}_1=\frac{1\left(1+1\right)}{2}\\ =1\end{array}$

Since the first term of the proposition is 1. So $S_n$ is valid for $n=1$ .

Assume that $S_n$ is valid for $n=k$ , i.e. $S_k=1+2+3+\mathrm{...}+k=\frac{k\left(k+1\right)}{2}$ .

Now verify that $S_n$ is valid for $n=k+1$ . Add $\left(k+1\right)$ to each side of $S_k$ as shown below.

  $\begin{array}{c}{S}_k+\left(k+1\right)=\frac{k\left(k+1\right)}{2}+\left(k+1\right)\\ 1+2+3+\mathrm{...}+k+\left(k+1\right)=\frac{k\left(k+1\right)}{2}+\frac{2\left(k+1\right)}{2}\\ =\frac{k^2+3k+2}{2}\\ =\frac{\left(k+1\right)\left(k+2\right)}{2}\end{array}$

Put the value of $n=k+1$ in $S_n$ as follows:

  $\begin{array}{c}{S}_n=\frac{\left(k+1\right)\left(k+2\right)}{2}\\ =S_{k+1}\end{array}$

Thus if the given proposition is valid for $n=k$ then it is also valid for $n=k+1$ . Since the given proposition is valid $n=1$ then it is also valid for $n=2,n=3$ and so on. Thus, the given statement is valid for all integers.