Question

if alpha and beta are the greatest divisors of n(n^2-1) and 2n(n^2+2) respectively for all n belongs to N then alpha x beta =​

Answer 1

Answer with explanation:

It is given that, α and β are greatest divisors of $n(n^2-1) {\text{and}} 2n(n^2+2)$ respectively.

For, n=1, value of , $n(n^2-1)=1 \times (1^2-1)=1 \times 0=0$

So, →we will start from, n=2.

$for, n=2, n(n^2-1)=2\times (2^2-1)\\\\ 2 \times 3=6\\\\ \text{Similarly for}}, n=2,2n(n^2+2)=2\times 2\times(2^2+2)=4\times (4+2)\\\\4 \times 6=24$

Greatest Factor of , 6 is α= 3

Greatest factor of ,24 is β= 12

So, αβ=3 ×12=36

→Now, if we take, n=3,then

$for, n=3, n(n^2-1)=3\times (3^2-1)\\\\ 3 \times 8=24\\\\ \text{Similarly for}}, n=3,2n(n^2+2)=2\times 3\times(3^2+2)=6\times (9+2)\\\\6 \times 11=66$

Greatest Factor of , 24 is α= 12

Greatest factor of ,66 is β= 33

So, αβ=12 × 33=396

⇒Looking at the options, n=2 , gives the correct Answer.

So,⇒ αβ=36

Option 2