Is a number member of sequence?

[cfg]

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!

 

[eddybob123]

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!

Is that $\frac{n^2}{2}-3n+\frac{5}{2}$?

If so, then if 10 is a member of the sequence, then there is a positive integer n that satisfies $$\frac{n^2}{2}-3n+\frac{5}{2}=10$$
Either find such a solution to the equation (solve for n) or prove that there isn't one. Do the same for 6.

 

[MarkFL]

While this may be beyond the scope of what is expected or even needed here, we could observe that since the closed form of the sequence is a quadratic with real coefficients, then the recursive form will come from the characteristic equation:

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

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 $a_n=10$?