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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.

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    $\begingroup$ Hint: Use the fact that $u_n = r^n$, try writting your equation using this fact $\endgroup$ – JoseSquare Feb 11 at 18:08
  • $\begingroup$ But isn't it $u_1 \cdot r^n$ ? $\endgroup$ – Daniel Oscar Feb 11 at 18:10
  • $\begingroup$ @DanielOscar You want $u_n = u_0 r^n$---notice that this formula gives an identity in the special case $n = 0$. $\endgroup$ – Travis Willse Feb 11 at 18:14
  • $\begingroup$ Doesn’t matter, look at the answer Dr.Sonnhard post $\endgroup$ – JoseSquare Feb 11 at 18:14
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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.

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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:

$u_{1+p}*u_{n-p} = r^p*u_1*r^{n-p-1}*u_1 = u_1*r^{n-1}*u_1 = u_1*u_n$.

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