Sequences & Series Question - None

Question

If $b = ar$, $c = ar^2$, and $d = ar^3$, then $(b - c)^2 + (c - a)^2 + (d - b)^2$ is equal to

Accepted Answer

Solution:

Given: $b = ar$, $c = ar^2$, $d = ar^3$

$(b - c)^2 + (c - a)^2 + (d - b)^2$

$= (ar - ar^2)^2 + (ar^2 - a)^2 + (ar^3 - ar)^2$

$= a^2r^2(1 - r)^2 + a^2(r^2 - 1)^2 + a^2r^2(r^2 - 1)^2$

$= a^2(1 - r)^2\{r^2 + (r + 1)^2 + r^2(r + 1)^2\}$

$= a^2(1 - r)^2(r^4 + 2r^3 + 3r^2 + 2r + 1)$

$= a^2(1 - r)^2(1 + r + r^2)^2$

$= a^2(1 - r^3)^2$

$= (a - ar^3)^2 = (a - d)^2$

Question: