Question

Confused on what do do first . Can anyone help ?

Answer 1

So solve for q

first factor q out of the summation$q \sum^{\inf}_{k=2} (\frac{2}{3})^k=8$

now, determine what the summation is
$\sum^{\inf}_{k=2} (\frac{2}{3})^k =?$

its been a while since ive done summations so i dont remember any tricks but that summation is essentially equal to
$\frac{2}{3} ^2 +\frac{2}{3}^3+\frac{2}{3}^4+\frac{2}{3}^5+...$

or factored to be something like this
$\frac{2}{3}^2 *(1+\frac{2}{3}+\frac{2}{3}^2+\frac{2}{3}^3+...)$
which i believe there's a formula for

regardless using a calculator, the summation turns out to be 4/3 i think

you should definitely double check this step

so replacing the summation for 4/3, the equation is now
q*(4/3) =8

pretty easy to solve from here
divide 4/3 to both sides to get q


any questions?