If $p + q = 1$, then $\sum_{r=0}^{n} r^3 {}^nC_r p^r q^{n-r} =$
$np(n^2 p - (n-1)p + 1)$
$np((n^2 - n)p^2 + 2(n-1)p + 1)$
$np((n^2 - 3n + 2)p^2 + 2(n-1)p + 1)$
$np((n^2 - 3n + 2)p^2 + 3(n-1)p + 1)$
Step-by-Step Solution
Key Concept: For an equilateral triangle with one vertex at the origin, the sum of squares equals the sum of pairwise products.
Given $z_1 + z_2 = -1$ and $z_1z_2 = \frac{b}{3}$, with $z_1, z_2$ and the origin forming an equilateral triangle, we use the identity $z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_2z_3 + z_3z_1$ where $z_3 = 0$. This gives $(z_1 + z_2)^2 - 2z_1z_2 = z_1z_2$, so $(−1)^2 = 3z_1z_2$, yielding $1 = b$.
Correct Answer: I need to find $\sum_{r=0}^{n} r^3 \binom{n}{r} p^r q^{n-r}$ where $p + q = 1$.
Let me use the binomial theorem approach. If we let $f(x) = (x+q)^n = \sum_{r=