For all n ∈ N, 3n^5 + 5n³ + 7n is divisible by: - Vidyadip

MCQ

For all $n ∈ N, 3n^5 + 5n³ + 7n$ is divisible by:

A
5
✓
15
C
10
D
3

✓

Answer

Correct option: B.

15

15

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

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