1

a, Is it true, that if $n \gt 1$ odd number, then we can express ${n^2-1\over 4}$ as multiply of two consecutive numbers?

b, Is it true, that we can express ${n^4+2n^3+3n^2+2n\over 4}$ in case of n= all natural numbers, as multiply of two consecutive numbers?

What I did already: I know, that the multiply of two consecutive numbers can end to 0,or 2, or 6.

a. ${n^2-1\over 4} = {(3+2m)^2-1\over 4}= {9+2*3*2m+4m^2-1\over 4} = {8+12m+4m^2\over 4}=2+3m+m^2$

m= 0,1,2,3...

Holdviola
  • 79
  • 4
    Just put $n=2m+1$ and see what happens to $\frac {n^{2}-1} 4$ – geetha290krm Oct 12 '22 at 12:26
  • 1
    In your work on a), you are almost done. You just need to factorize the polynomial in m. I might as well provide it here: $m^2+3m+2 = (m+1)(m+2)$.

    In (b), try to figure out a factorization as well of the numerator, and see if you can't prove that either one of the factors is divisible by 4, or both are divisible by 2. You might want to consider the cases of $n$ even and odd separately.

     – SomeCallMeTim Oct 12 '22 at 12:34

1 Answers1

1

a. $\frac{(2m+1)^2-1}{4}=m(m+1)$.

b. $\frac{n^4+2n^3+3n^2+2n}{4}=\left(\frac{n(n+1)}{2}\right)\left(\frac{n(n+1)}{2}+1\right)$

Bob Dobbs
  • 10,988