The value of \(\frac{1}{81^{n}}-\frac{10}{81^{n}}\) \({ }^{2 n} C_{1}+\frac{10^{2}}{81^{n}} \cdot{ }^{2 n} C_{2}+\frac{10^{3}}{81^{n}} \cdot{ }^{2 n} C_{3}\) + .... + \(\frac{10^{2 n}}{81^{n}}\) is equal to:

Binomial Theorem

Grade 11

Question:

The value of \(\frac{1}{81^{n}}-\frac{10}{81^{n}}\) \({ }^{2 n} C_{1}+\frac{10^{2}}{81^{n}} \cdot{ }^{2 n} C_{2}+\frac{10^{3}}{81^{n}} \cdot{ }^{2 n} C_{3}\) + .... + \(\frac{10^{2 n}}{81^{n}}\) is equal to:

1

2

\(\frac{1}{2}\)

0

Step-by-Step Solution

Key Concept: The expression is a binomial expansion of (1-10)^{2n} scaled by 1/81^n, which simplifies to 1 because (-9)^{2n} is equal to 81^n.

1

Correct Answer: A

Question:

Step-by-Step Solution

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