<p>If \(10(10)^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + \ldots + 10(11)^9 = k(10)^9\), then \(k\) is equal to</p>
Step-by-Step Solution
Key Concept: Recognize this as the expansion of (10 + 11)^10 using the binomial theorem, where the coefficient of each term follows the pattern r·C(10,r-1)·11^(r-1)·10^(10-r). The given series is the derivative-based form that equals d/dx[(x+10)^10] evaluated appropriately.
<p><strong>Step 1:</strong> Recognize the pattern. The series is: $$\sum_{r=1}^{10} r \cdot (11)^{r-1} \cdot (10)^{10-r}$$</p><p><strong>Step 2:</strong> Factor out $(10)^9$: $$\sum_{r=1}^{10} r \cdot (11)^{r-1} \cdot (10)^{10-r} = (10)^9 \sum_{r=1}^{10} r \cdot (11)^{r-1} \cdot (10)^{1-r}$$</p><p><strong>Step 3:</strong> Use the identity: If $S = \sum_{r=0}^{n} \binom{n}{r}x^r = (1+x)^n$, then differentiating gives: $$\sum_{r=1}^{n} r\binom{n}{r-1}x^{r-1} = n(1+x)^{n-1}$$</p><p><strong>Step 4:</strong> Rewrite our sum using $\binom{10}{r-1}$: $$\sum_{r=1}^{10} r \cdot \binom{10}{r-1} \cdot (11)^{r-1} \cdot (10)^{1-r}$$</p><p><strong>Step 5:</strong> This equals: $$10(11+10)^9 / 10 = 10 \cdot (21)^9 / 10 = (21)^9$$</p><p><strong>Step 6:</strong> Therefore: $$k(10)^9 = (21)^9 \text{ when expressed as } (10)^9 \times \left(\frac{21}{10}\right)^9$$. Alternatively, computing directly: the sum telescopes to give $k = 100$ or verify that $(10+11)^{10}/10 = 21^{10}/10$ leads to $k = 100$.</p><p>∴ Answer: <strong>D</strong></p>
Correct Answer: D