The sum to infinity of a geometric progression is 6. The sum to infinity of the squares of each term in the progression is 12. Find the sum to infinity of the cubes of each term in the progression. A 8 B 18 C 24 D 216/7 E 72 F 216
Lets say a = first term and r = common ratio. I understand how the mark scheme derived their answer but I am unsure as to what I did incorrect in my calculation.
So I got:
$$\frac{a}{1-r} = 6$$ and $$\frac{a}{1-r^2} = 12$$
$6(1-r) = 12(1-r^2)$
$12r^2 - 6r - 6 = 0$
$2r^2 - r - 1 = 0$
$(2r + 1)(r - 1) = 0$
because |r| < 1
$$\therefore r = -1/2$$
so $a = 6 - 6(\frac{-1}{2})$
$$a = 9$$
The sum to infinity of cubes of each term: $$\frac{a}{1-r^3}$$
$$\frac{9}{1 - (\frac {-1}{2})^3} = \frac {9}{1 + (\frac{1}{8})} = 8$$
However the answer is 216/7. I am unsure as to where I went wrong in my calculation. I am guessing I shouldn't have multiplied out to get $a = 12(1 - r^2)$ because r could take two values? One being the correct value and the other being the opposite sign of the correct value?
The original series is $a+ar+ar^2+ar^3+\cdots$ and its sum is indeed $\frac a{1-r}$. But the series of the squares is $a^2+a^2r^2+a^2r^4+a^2r^6+\cdots$ and its sum is $\frac{a^2}{1-r^2}$.