Question

The value of $\frac{1}{81^{n}}$–$\frac{10}{81^{n}}$²ⁿC₁ +$\frac{10^{2}}{81^{n}}$²ⁿC₂ – $\frac{10^{3}}{81^{n}}$²ⁿC₃ +....+ $\frac{10^{2n}}{81^{n}}$ is –

• A. 2
• B. 0
• C. ½
• D. 1

Answer 1

Answer: (D)

1

Answer 2

$\frac{1}{81^{n}}$–$\frac{10}{81^{n}}$²ⁿC₁ + $\frac{10^{2}}{81^{n}}$²ⁿC₂ – $\frac{10^{3}}{81^{n}}$ ²ⁿC₃ + ..... + $\frac{10^{2n}}{81^{n}}$

= $\frac{1}{81^{n}}$ [²ⁿC₀ – ²ⁿC₁ 10 + ²ⁿC₂ 10² – ²ⁿC₃ 10³ + ....... + ²ⁿC_2n 10²ⁿ]

= $\frac{1}{81^{n}}$ (1 – 10)²ⁿ = $\frac{( - 9)^{2n}}{(9^{2})^{n}}$ = $\frac{( - 1)^{2n}9^{2n}}{9^{2n}}$= 1