I need assistance with the following question:
On geometric sequence $q=2$ and the sum of the sequence is 30. The sum of $(a_1a_2+a_2a_3+a_3a_4+...)$ is $680$ how much terms on the sequence ?
(q is the factor between the terms, and n is the amount of terms on the sequence)
My try:
*$S_n=a_1(2^n-1)=30$
**$(a_1a_2+a_2a_3+a_3a_4+...)=680 >>\frac{a_1^2q((q^2)^{n-1}-1)}{q^2-1}=680>> q=2 >>$ $\frac{a_1^22((2^2)^{n-1}-1)}{2^2-1}=680 >> 2a_1^2(2^{2n-2}-1)=2040$
$a_1^2(2^{2n-2}-1)=1020$
then I pluged in $a_1$ from * and call $2^n=t$
$\frac{900(t^2\frac{1}{4}-1)}{t^2-2t+1}=1020$
but then I got complex answers here.
Thanks!
If $a_n = qr^n$ then $a_na_{n+1} =qr^nqr^{n+1} =q^2r^{2n+1} =q^2rr^{2n} =q^2r(r^2)^{n} $.
The sums $s(m) =\sum_{n=0}^{m-1} a_n =\sum_{n=0}^{m-1} qr^n =q\dfrac{r^m-1}{r-1} =30 $ and $t(m) =\sum_{n=0}^{m-1} a_na_{n+1} =\sum_{n=0}^{m-1} q^2r(r^2)^{n} =q^2r\sum_{n=0}^{m-1} (r^2)^{n} =q^2r\dfrac{(r^2)^m-1}{r^2-1} =680 $.
Replace the numbers by $u=30$ and $v=680$ so $u = q\dfrac{r^m-1}{r-1} $ and $v =q^2r\dfrac{(r^2)^m-1}{r^2-1}$.
Now we have to solve these for $r$ (or $q$) and $m$.
If we know $q$ then, dividing, $\dfrac{v}{u} =qr\dfrac{(r^2)^m-1}{r^2-1}\dfrac{r-1}{r^m-1} =qr\dfrac{r^m+1}{r+1} $ so $r^m =\dfrac{v(r+1)}{uqr}-1 $ and $r^m =\dfrac{u(r-1)}{q}+1 $.
Therefore $\dfrac{u(r-1)}{q}+1 =\dfrac{v(r+1)}{uqr}-1 $. Multiplying by $uqr$, $u^2r(r-1)+2uqr = v(r+1)$ or $u^2r^2+(2uq-v)r-v =0$. This is a quadratic that can be solved for $q$. This, in turn, will give $r^m$ and therefore $m$.
If we know $r$, from $u = q\dfrac{r^m-1}{r-1} $ and $v =q^2r\dfrac{(r^2)^m-1}{r^2-1}$, squaring the first gives $u^2 = q^2\dfrac{(r^m-1)^2}{(r-1)^2} $.
Dividing gives
$\begin{array}\\ \dfrac{v}{u^2} &=\dfrac{q^2r\dfrac{(r^2)^m-1}{r^2-1}}{q^2\dfrac{(r^m-1)^2}{(r-1)^2}}\\ &=\dfrac{r(r-1)^2((r^2)^m-1)}{(r^2-1)(r^m-1)^2}\\ &=\dfrac{r(r-1)(r^m+1)}{(r+1)(r^m-1)}\\ \end{array} $
so $\dfrac{v}{u^2}(r+1)(r^m-1) =r(r-1)(r^m+1) $ or $r^m(\dfrac{v}{u^2}(r+1)-r(r-1)) =r(r-1)+\dfrac{v}{u^2}(r+1) $ or $r^m =\dfrac{r(r-1)+\dfrac{v}{u^2}(r+1)}{\dfrac{v}{u^2}(r+1)-r(r-1)} $.
This will give $m$ and $q$ then follows.