For all n N, 3n^5+ 5n^3+ 7n is divisible by: - Vidyadip

MCQ

For all $n \in N, 3n^5+ 5n^3+ 7n$ is divisible by:

A
$5$
✓
$15$
C
$10$
D
$3$

✓

Answer

Correct option: B.

$15$

Given number $=3 n^5+5 n^2+7 n $
Let $ n=1,2,3,4,...... $
$ 3 n^5+5 n^3+7 n$
$=3 \times 1^2+5 \times 1^3+7 \times 1$
$=3+5+7$
$=15 $
$ 3 n^5+5 n^3+7 n$
$=3 \times 2^5+5 \times 2^3+7 \times 2$
$=3 \times 32+5 \times 8+7 \times 2$
$=96+40+14$
$=15 \times 10$
$=150$
$ 3 n^5+5 n^3+7 n$
$=3 \times 3^5+5 \times 3^3+7 \times 3$
$=3 \times 243+5 \times 27+7 \times 3$
$=729+135+21$
$=15 \times 59$
$=885$
Since, all these numbers are divisible by $15$ for $n = 1, 2, 3, …..$
So, the given number is divisible by $15.$

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