- Thread starter
- #1

- Thread starter cfg
- Start date

- Thread starter
- #1

- Aug 18, 2013

- 76

Is that $\frac{n^2}{2}-3n+\frac{5}{2}$?an(n in subindex)=(1/2)*n^2-3n+5/2, when n ≥1

Is number 10 member of that sequence? what about number 6?Create equation to solve it.

If someone can help with this problem please, it will be much appreciated!

If so, then if 10

Either find such a solution to the equation (solve for n) or prove that there isn't one. Do the same for 6.

- Admin
- #3

\(\displaystyle (r-1)^3=r^3-3r^2+3r-1\)

and so the sequence may be defined recursively as:

\(\displaystyle a_{n+3}=3a_{n+2}-3a_{n+1}+a_{n}\)

where:

\(\displaystyle a_1=0,\,a_2=-\frac{3}{2},\,a_3=-2\)

Another thing we might look at is the equation:

\(\displaystyle \frac{n^2-6n+5}{2}=a_n\)

\(\displaystyle n^2-6n+5-2a_n=0\)

If there is going to be an integral root, then the discriminant will be a perfect square, the square of an even number:

\(\displaystyle (-6)^2-4(1)\left(5-2a_n \right)=(2m)^2\) where \(\displaystyle 0\le m\in\mathbb{Z}\)

\(\displaystyle 9+2a_n-5=m^2\)

\(\displaystyle 4+2a_n=m^2\)

\(\displaystyle 2\left(2+a_n \right)=m^2\)

So, we can easily see that when \(\displaystyle a_n=6\implies m=4\). What about when $a_n=10$?