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- #1
is a number member of sequence?
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eddybob123
Active member
- 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 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.
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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 $a_n=10$?
\(\displaystyle (r-1)^3=r^3-3r^2+3r-1\)
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 $a_n=10$?