Binomial Theorem
Binomial Series Summation
Grade 11
Question:
<p>Find the value of <span>\(\dfrac{1}{81^n} - \dfrac{10}{81^n}\,{}^{2n}C_1 + \dfrac{10^2}{81^n}\,{}^{2n}C_2 - \dfrac{10^3}{81^n}\,{}^{2n}C_3 + \cdots + \dfrac{10^{2n}}{81^n}\)</span>.</p>
Step-by-Step Solution
Key Concept: Recognize the series as the binomial expansion of (1 - x)^(2n) where x = 10/81, then substitute x = 1 to collapse the alternating series to a single term.
<p><strong>Step 1:</strong> Identify the series structure. The given expression is:</p><p>$$\sum_{r=0}^{2n} (-1)^r \binom{2n}{r} \frac{10^r}{81^n}$$</p><p><strong>Step 2:</strong> Factor out $\frac{1}{81^n}$ and rewrite:</p><p>$$\frac{1}{81^n} \sum_{r=0}^{2n} \binom{2n}{r} (-10)^r$$</p><p><strong>Step 3:</strong> Recognize this as the binomial expansion of $(1 + x)^{2n}$ where $x = -10$:</p><p>$$(1-10)^{2n} = \sum_{r=0}^{2n} \binom{2n}{r}(1)^{2n-r}(-10)^r$$</p><p><strong>Step 4:</strong> Substitute and evaluate:</p><p>$$\frac{1}{81^n}(1-10)^{2n} = \frac{(-9)^{2n}}{81^n} = \frac{9^{2n}}{81^n}$$</p><p><strong>Step 5:</strong> Simplify using $81 = 3^4$:</p><p>$$\frac{(3^2)^{2n}}{(3^4)^n} = \frac{3^{4n}}{3^{4n}} = 1$$</p><p>∴ Answer: <strong>1</strong></p>
Correct Answer: 1