Question

Prove that the sum of the first n nonnegative integers is $\frac{\mathbf{n}\left(n+1\right)}{\mathbf{2}}$. [Hint: Let P(k) be the statement:$\mathbf{0}\mathbf{+}\mathbf{1}\mathbf{+}\mathbf{2}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}\mathbf{k}\mathbf{=}\frac{\mathbf{k}\left(k+1\right)}{\mathbf{2}}$ .

Short answer

The sum of the first n nonnegative integers is $\frac{\mathit{n}\mathit{(}\mathit{n}\mathit{+}\mathit{1}\mathit{)}}{\mathit{2}}$.

Step-by-step solution

Step 1

Consider the following equation for P(k):

$0+1+2+....+\mathrm{k}=\frac{k(\mathrm{k}+1)}{2}$

Here, k is any real number.

Step 2

Substitute 1 for k in the equation $0+1+2+....+\mathrm{k}=\frac{k(\mathrm{k}+1)}{2}$ as follows;

$$\begin{gathered} \frac{1\left(1+1\right)}{2}=\frac{1\left(2\right)}{2} \\ =\frac{2}{2} \\ =1 \end{gathered}$$

Step 3

So, the equation is true for some $k=n+1$, that is;

$0+1+2+....+n=\frac{n(n+1)}{2}$

Hence, the expression is verified.