Question

The sum $S_n=n^3+3n^2+5n+3$ is divisible by

• A. $3\forall n\in N$
• B. $4\forall n\in N$
• C. $5\forall n\in N$
• D. can’t be determined

Answer 1

Answer: (A)

Let $P\left(n\right)$ be the statement given by
$P\left(n\right):S_n=n^3+3n^2+5n+3$ is divisible by $3$
We have, $P\left(1\right):S_1=1^3+3{\left(1\right)}^2+5\left(1\right)+3=12$, which is divisible by $3$
$\therefore P\left(1\right)$ is true.
Let $P\left(k\right)$ be true. Then,
$P\left(k\right):S_k=k^3+3k^2+5k+3$ is divisible by $3$
$\Rightarrow S_k=k^3+3k^2+5k+3=3\lambda$, for some $\lambda \in N...\left(i\right)$
We now wish to show that $P\left(k+1\right)$ is true, i.e.
${\left(k+1\right)}^3+3{\left(k+1\right)}^2+5\left(k+1\right)+3$ is divisible by $3$
Now, ${\left(k+1\right)}^3+3{\left(k+1\right)}^2+5\left(k+1\right)+3$
$=\left(k^3+3k^2+5k+3\right)+3k^2+9k+9$
$=3\lambda +3\left(k^2+3k+3\right)$ [Using $\left(i\right)$]
$=3\left[\lambda +k^2+3k+3\right]$, which is divisible by $3$
Thus, $P\left(k+1\right)$ is true, whenever $P\left(k\right)$ is true.
Hence, by the principle of mathematical induction, the statement is true for all $n\in N$.