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?
