Question 13 Marks
If $P(n)$ is the statement " $n^2+n$ is even", and if $P(r)$ is true, then $P(r+1)$ is true.
Answer
View full question & answer→$P(n): n^2+n$ is even
Given, $P(r)$ is true
$\Rightarrow r^2+r$ is even
$\Rightarrow r ^2+ r =2 \lambda \ldots$ (1)
Now,
$(r+1)^2+(r+1)$
$=r^2+1+2 r+r+1$
$=(r+1)^2+2 r+2$
$=2 \lambda+2 r+2[\text { Using equation (1)] }$
$=2(\lambda+ r +1)$
$=2 \lambda$
$\Rightarrow(r+1)^2+(r+1) \text { is even }$
$\Rightarrow P(r+1) \text { is true }$
Given, $P(r)$ is true
$\Rightarrow r^2+r$ is even
$\Rightarrow r ^2+ r =2 \lambda \ldots$ (1)
Now,
$(r+1)^2+(r+1)$
$=r^2+1+2 r+r+1$
$=(r+1)^2+2 r+2$
$=2 \lambda+2 r+2[\text { Using equation (1)] }$
$=2(\lambda+ r +1)$
$=2 \lambda$
$\Rightarrow(r+1)^2+(r+1) \text { is even }$
$\Rightarrow P(r+1) \text { is true }$