The terms of an infinite series S are formed by adding together the corresponding terms in two infinite geometric series, T and U.
The first term of T and the first term of U are each 4.
In order, the first three terms of the combined series S are 8, 3, and 5/4.
What is the sum to infinity of S?
$T_1 = 4, U_1 = 4$
$T_2 + U_2 =3$
$T_3 + U_3 = 5/4$
therefore
$4r_t + 4r_u = 3$
$4r_t^2 + 4r_u^2 = 5/4$
How can I continue from there? When I try to solve by replacing $r_u$ with $\frac{3-4r_t}{4}$ I find two values for $r_t$ and I don't know which one is the correct one.
Any help is very appreciated!
Best Answer
We have $$r_t + r_u = 3/4$$ $$r_t^2+r_u^2=5/16$$ Squaring the first equation gives $$r_t^2+2r_tr_u+r_u^2=9/16$$ Then subtracting the second equation from this gives $$2r_tr_u=1/4\qquad r_tr_u=1/8$$ Thus, the polynomial $r^2-\frac34r+\frac18$ has roots $r_t,r_u$ by Viète's formulas. We get that the roots are $\frac14$ and $\frac12$. The sum to infinity of $S$ is then the sum to infinity of its components: $$\frac4{1-1/4}+\frac4{1-1/2}=\frac{16}3+8=\frac{40}3$$