Given a sequence ($u_1; u_2; u_3;...; u_n$), $n\in \mathbb{N}$, of $n$ terms in geometric series, show that for every natural number $0\le p\le n-1$,
$$u_{1+p} \cdot u_{n-p}=u_1 \cdot u_n$$
I know I'm not supposed to ask for solutions for textbooks exercises, but I'm teaching myself math without ever have had a math class in highschool. I would appreciate not only a solution but an intuitive description.
We will use the fact, that $$u_n=u_1q^{n-1}$$ so $$u_{n-p}=u_1q^p$$ and $$u_{n-p}=u_1q^{n-p-1}$$ and we get $$u_{1+p}\cdot u_{n-p}=u_1^2q^{n-1}$$ and the right-hand side: $$u_1\cdot u_n=u_1^2q^{n-1}$$ and this is the same term.
The definition of a geometric sequence is that $u_{k+1} = r*u_{k}$ for a constant $r$.
So $u_{1+p} * u_{n-p} = $
$(u_p*r)*u_{n_p} = u_p*(r*u_{n-p})=$
$u_p*u_{n-p+1} = $
$(u_{p-1}*r)*u_{n-p + 1} = u_{p-1}*(r*u_{n-p + 1}) =$
$u_{p-1}*u_{n-p + 2}= $
...
$u_{p-k}*u_{n-p + k+1}=$
...
$u_2*u_{n-1}=$
$u_1*r*u_{n-1} =$
$u_1*u_n$.
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It's worth noting Via induction $u_1$ (the first term) is some constant.
$u_2 = r*u_1$ and $u_3 = r*u_2 = r*(r*u_1) = r^2*u_1$ and $u_4= r*u_3 = r(r^2*u_1)=r^3*u_1$ and inductively so on: $u_{k+1} = r^k*u_1$.
Which in turn means:
$r^k*u_m = r^k*(r^{m-1}*u_1) = r^{k+m-1}*u_1 = u_{k+m}$.
So:
$u_{1+p}*u_{n-p} = r^p*u_1*r^{n-p-1}*u_1 = u_1*r^{n-1}*u_1 = u_1*u_n$.
