# Prove that given a geometric series, $u_{1+p} \cdot u_{n-p}=u_1 \cdot u_n$

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.

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

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:

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