Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

The value of (1 / 81n) - (10 / 81n) 2nC1 + (102 / 81n) 2nC2 - (103 / 81n) 2nC3 + ….. + (102n / 81n) is

1) 2

2) 0

3) 1/2

4) 1

Solution: (4) 1

\(\begin{array}{l}\frac{1}{81^{n}}-\frac{10}{81^{n}} 2_{n_{C_{1}}}+\frac{10^{2}}{81^{n}} 2 n_{C_{2}}-\frac{10^{3}}{81^{n}} 2 n_{C_{3}}+\cdots+\frac{10^{2 n}}{81^{n}} \\ =\frac{1}{81^{n}}\left[1-102 n_{C_{1}}+10^{2} 2 n_{C_{2}}-10^{3} 2 n_{C_{3}}+\cdots+10^{2 n}\right]\\ =\frac{1}{81^{n}}(1-10)^{2 n} \\ =\frac{9^{2 n}}{81^{n}}\\=1\end{array} \)

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