Question

the product of any three consecutive numbers is always divisible by 6 give an example and justify your answer​

Answer 1

Here is a formal proof with all the steps very explicitly outlined. The proper statement of this theorem is as follows:

For every natural number k greater than or equal to 1, there exists such a natural number a that:

k(k+1)(k+2)=6a

This can be proved by induction, for k=1:

1(1+1)(1+2)=6a

6=6a

a=1

For k=n+1 the theorem states:

(n+1)(n+2)(n+3)=6a

To use the induction hypothesis, expand the multiplication using the last bracket:

(n+1)(n+2)(n+3)=

(n+1)(n+2)n+(n+1)(n+2)3=

n(n+1)(n+2)+3(n+1)(n+2)

By the induction hypothesis, n(n+1)(n+2) is equal to 6b for some natural number b. (n+1)(n+2) is a product of an odd number and even number, hence a number of the form 2c for some other natural number c:

n(n+1)(n+2)=6b

(n+1)(n+2)=2c

Now just substitute this into the previous expansion:

(n+1)(n+2)(n+3)=

n(n+1)(n+2)+3(n+1)(n+2)=

6b+3*2c=

6b+6c=

6(b+c)

a=b+c

So, finally:

(n+1)(n+2)(n+3) = 6[(n(n+1)(n+2))/6+((n+1)(n+2))/2]

With (n(n+1)(n+2))/6+((n+1)(n+2))/2 being a natural number, which was to be proven. This proof also illustrates the fact that, having a triple of consecutive numbers and their product, adding to this product the product of the last two numbers and the number 3 results in the product of the next three consecutive numbers,

e.g.: (by u r demand❤️)

1*2*3=6

2*3*4=24=6+3*(2*3)

3*4*5=60=24+3*(3*4)

4*5*6=120=60+3*(4*5)

5*6*7=210=120+3*(5*6)

...so on....

Follow me dear for your future doubts ❤️

.