Question

Evaluate the following expression: \[ \frac{1}{81^n} - \binom{2n}{1} . \frac{10}{81^n} + \binom{2n}{2} . \frac{10^2}{81^n} - .s + \frac{10^{2n}}{81^n} = ? \]

• A. \( 0 \)
• B. \( (-1)^n \)
• C. \( 1 \)
• D. \( 81 \)

Hint:

Look for patterns in binomial expansions and try to simplify using known formulas like \((a + b)^n\).

Answer 1

The Correct Option is C

Step 1: Recognize the binomial expansion: \[ \left( \frac{10 - 1}{9} \right)^{2n} = \left( \frac{9}{9} \right)^{2n} = 1 \] Step 2: Let us write the expression: \[ \sum_{k=0}^{2n} (-1)^k \binom{2n}{k} . \frac{10^k}{81^n} = \frac{1}{81^n} \sum_{k=0}^{2n} \binom{2n}{k} (-10)^k \] Step 3: Use Binomial Theorem: \[ \sum_{k=0}^{2n} \binom{2n}{k} (-10)^k = (1 - 10)^{2n} = (-9)^{2n} = 81^n \] Step 4: Final result: \[ \frac{81^n}{81^n} = 1 \]