Question

Prove that the sum of 3 consecutive odd numbers is always a multiple of 3.​

Answer 1

Answer:

sum of 3 consecutive odd numbers is always a multiple of 3.

Step-by-step explanation:

You pick some odd number n .

Then, you take the two odd numbers either side of it. They will be n−2 and n+2 .

If we sum those 3 consecutive odd numbers, we get:

(n−2)+(n)+(n+2)

=n+n+n

=3n

Therefore, the sum of 3 consecutive odd numbers, or indeed any 3 consecutive even numbers equal to 3 times the middle number.

Since that’s the case, it is also the case that it is divisible by 3.

Answer 2

let's name any odd number "n"

every second number is an odd number.

so the three nearest odd numbers are going to be:

n, n+2, n+4

sum of them will be n+(n+2)+(n+4) which is 3n+6.

when we divide 3n+6 by 3 we'll get $\frac{3n+6}{3}$ = n+2.

n+2 is a whole number.