Question

$\forall n\in N;3n^5+5n^3+7n$ is divisible by

• A. $3$
• B. $5$
• C. $10$
• D. $15$

Answer 1

Answer: (A), (B), (D)

$3$
$5$
$15$

Put $n=1$ in the given expression
$3n^5+5n^3+7n$
$3+5+7=15$
which is divisible by $3,5,15$
Put $n=2$
$3\left(32\right)+5\left(8\right)+14=150$
which is divisible by $3,5,10,15$
Since, P(1) and P(2) both are divisible by $3,5,15$
Every number divisible by 15 is divisible by 3 and 5 also.
So, let us assume $P\left(n\right)=3n^5+5n^3+7n$ is divisible by 15
$\Rightarrow 3n^5+5n^3+7n=15\lambda$
Now, we have to prove P(n+1) is true
Consider, $P(n+1)=3(n+1{)}^5+5(n+1{)}^3+7(n+1)$
$=3(n^5+5n^4+10n^3+10n^2+5n+1)+5(n^3+1+3n^2+3n)+7n+7$
$=(3n^5+5n^3+7n)+15n^4+30n^3+45n^2+30n+15$
$=15\lambda +15(n^4+2n^3+3n^2+2n+1)$
$=15(\lambda +n^4+2n^3+3n^2+2n+1)$
which is divisible by $15.$
Hence, $P(n+1)$ is divisible by $15.$
Hence, by mathematical induction, given expression is divisible by 15, for all $n\in N$
Hence, given expression is divisible by $3,5,15$