Question

In Problem, use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers $n$.

role="math" localid="1647672162255" $4+3+2+...+(5-n)=\frac{1}{2}n(9-n)$

Short answer

It is shown that the statement$4+3+2+...+(5-n)=\frac{1}{2}n(9-n)$ is true for all natural numbers

Step-by-step solution

Step 1. Given information

The statement is$4+3+2+...+(5-n)=\frac{1}{2}n(9-n)$.

Step 2. Show that the given statement is true for all natural numbers.

Put $n=1$in the given statement.

$\frac{1}{2}1(9-1)=4$

Assume that the statement is true for $n=k$

role="math" $4+3+2+\dots \dots \dots +(5-k)=\frac{1}{2}k(9-k)$

Now, find the sum for $k+1$terms.

role="math" localid="1647672985508" $\begin{array}{rcl}4+3+2+\dots \dots \dots +(5-k)+(5-(k+1\left)\right)& =& \frac{1}{2}k(9-k)+(4-k)\\ & =& \frac{1}{2}\left(9k-k^2\right)+\frac{1}{2}(8-2k)\\ & =& \frac{1}{2}\left(8+7k-k^2\right)\\ & =& \frac{1}{2}(k+1)(9-1-k)\\ & =& \frac{1}{2}(k+1)[9-(k+1\left)\right]\end{array}$

Thus, the statement is true for$k+1$

Since both the conditions of the Principle of Mathematical Induction are satisfied, the statement is true for all natural numbers.