Question

If r is an integer show that r(R^2-1)(3R+2) is always divisible by 24

Answer 1

Let f(r) = r(r² - 1)(3r + 2)
Can we rearrange it with proper way,
Let's do it, f( r) = r(r - 1)(r + 1)(r + 2 + 2r)
f(r) = (r - 1)r(r + 1)(r + 2) + (r - 1)r(r + 1)2r
Let, h(r) = (r - 1)r(r + 1)(r + 2) and p(r) = (r - 1)r(r + 1)2r

We can write h(r) = 4! × $^{(r+2)}C_4$
[ Because 4! $^{(r+2)}C_4$ = 4! × (r + 2)!/4!(r -2)! = (r+2)(r+1)r(r-1) ]
so, h(r) = 4! $^{(r+2)}C_4$ = 24 $^{(r+2)}C_4$
∴ h(r) is divisible by 24

we can also write p(r) = 2r × 3! × $^{(r+1)}C_3$
[ Because 3! $^{(r+1)}C_3$ = 3! × (r + 1)!/3!(r -2)! = (r +1)r(r-1)]
P(r) = 12r $^{(r+1)}C_3$
Now, case 1 :- when r is even number
then, p(r) must be divisible by 24
Case 2 :- when r is odd number ,
Then, $^{(r+1)}C_3$ = (r + 1)r(r -1)/6 , it's also even number
[ Because if we take r = 3 , (r + 1)r(r -1)/6 = 24/6 = 4 is an even number]
So, if r is odd then also P(r) is divisible by 24

Finally , f(r) = h(r) + p(r)
h(r) and p(r) both are divisible by 24
so, f(r) is divisible by 24

Answer 2

Answer: Let  r(r² - 1)(3r + 2)

                     = r(r - 1)(r + 1)(r + 2 + 2r)

                     = (r - 1)r(r + 1) (r+1)2 + 2

                     =(r - 1)r(r + 1) (r+1+1)2

                     =(r - 1)r(r + 1) (r+2)(r+2)

as the product of any 4 concecutive no is divisible by 24 it is also divisible by 24