Question

If n is any even number, then
n
(
n
2
+
20
)
is always divisible by

​

Answer 1

Answer:

It has to be proven that for all even positive integers n(n^2 + 20) is divisible by 48.

48 = 16*3 = 2^4*3

As n is an even positive integer, it can be written as 2*k where k is a positive integer.

n(n^2 + 20)

= 2k*(4k^2 + 20)

= 4*2*k(k^2 + 5)

= 8*k(k^2 - 1 + 6)

= 8k*(k^2 - 1) + 48k

= 8k(k - 1)(k + 1) + 48k

(k - 1)*k*(k + 1) is the product of three consecutive integers and therefore is divisible by 6.

= 8*6*m + 48k, where m and k are integers

= 48m + 48k

As both the terms have 48 as factors, the sum is divisible by 48

This proves that for all even positive integers n(n^2 + 20) is divisible by 48