MCQ
$n^2+ 3n$ is always divisible by which number, provided $n$ is an integer?
  • $2$
  • B
    $3$
  • C
    $4$
  • D
    $5$

Answer

Correct option: A.
$2$
$P(n) = n^2+ 3n$
$P(1) = 1 + 3$
$P(1) = 4$
Let’s assume that $P(k)$ is true and divisible by $4.$
Therefore, $P(k) = k^2+ 3k$ can be written as $4c.$
We need to check if $P(k + 1)$ is divisible by $4$
$P(k+1)=(k+1)^2+3(k+1) $
$P(k+1)=k^2+1+2 k+3 k+3 $
$P(k+1)=k^2+5 k+4 $
$P(k+1)=\left(k^2+3 k\right)+2 k+4 $
$P(k+1) = 4c + 2k + 4$
$P(k+1) = 4c + 2(k + 2)$
Clearly the second part of the equation is not divisible by $4.$
However $P(k) = 4c$ is divisible by $2$ and
$P(k + 1)$ is also divisible by $2.$
Therefore, $2$ divides $P(n)$

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