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How do we find the general form of $nth$ term of a sequence like this type of sequence :

$$ 1,2,6,24,120,...$$

so it's :

$$ ×2,×3,×4,...$$

which is :

$$+1,+1,+1,+1,...$$

I mean :

Do someone know any general form for a sequence like $$ × x, ×y, ×z,... $$ such that the difference between $x ; y$, $y; z$ are constant ;

or like $$+x, +y, +z..$$ such that $y/x$, $z/y..$ are constant?

Elie Makdissi
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    $1 = 1!$, $2 = 1 \cdot 2 = 2!$, $6 = 2! \cdot 3 = 3!$, $\cdots$ – Jay Nov 12 '21 at 18:28
  • thanks; what about the second part of question related to the general form? – Elie Makdissi Nov 12 '21 at 18:30
  • The first term is $1!$. The second term is $2!$. The third term is $3!$. Do you see a pattern? – Jay Nov 12 '21 at 18:32
  • The second part of your question is about arithmetic and geometric sequences. See https://bit.ly/3F2k2ls and https://www.mathsisfun.com/algebra/sequences-sums-geometric.html . – Michael Cohen Nov 12 '21 at 18:35

1 Answers1

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First Part

Was answered here https://math.stackexchange.com/a/4304279/772616

Second part

If $y-x=z-y=c$ where $c$ is some constant, then the best thing we could do is:

Suppose we have the sequence $u_n$, and $\frac{u_1}{u_0} = x$, $\frac{u_2}{u_1} = y = x + c$, $\frac{u_3}{u_2} = z = x + 2c$,... Then

$$ \prod_{k=1}^n \frac{u_k}{u_{k-1}} = x(x+d)(x+2d)...(x+(n-1)c)\therefore u_n=u_0\prod_{k=0}^{n-1}(x-ck) $$

Third Part

If $\frac{y}{x} = \frac{z}{y} = c$, then do the same thing as above.

aliberro
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