Question

By mathematical induction $p^{n+1}+(p+1{)}^{2n-1}$ is divisible by

• A. $p^2+p+1$
• B. $p^2+1$
• C. $p+1$
• D. None of these

Answer 1

The correct option is A $p^2+p+1$

Let $f\left(n\right)=p^{n+1}+{(p+1)}^{2n-1}$

we have $f\left(1\right)=p^2+p+1$ which is divisible by $p^2+p+1$

Now, assume that $f\left(m\right)$ is divisible by $p^2+p+1$

$\therefore p^{m+1}+{(p+1)}^{2m-1}=k(p^2+p+1)$ ...(1)

$f(m+1)=p^{m+2}+{(p+1)}^{2m+2-1}$

$=p^{m+2}+{(p+1)}^{2m-1}.{(p+1)}^2$

$=p^{m+2}+[k(p^2+p+1)-p^{m+1}]{(p+1)}^2$

$=p^{m+2}-{(p+1)}^2{p}^{m+1}+k{(p+1)}^2(p^2+p+1)$

$=p^{m+1}(p-p^2-2p-1)+k{(p+1)}^2(p^2+p+1)$

$=-(p^2+p+1)[-k{(p+1)}^2+p^{m+1}]$

Hence, $f(m+1)$ is divisible by $p^2+p+1$

Hence by mathematical induction $f\left(n\right)$ is divisible by $p^2+p+1$ for all $n\in N$