Solveeit Logo

Question

Mathematics Question on Binomial theorem

The value of 181n1081n2nC1+10281n2nC210381n2nC3+.....+102n81nis\frac{1}{81^n}-\frac{10}{81^n}\,^{2n}C_1+\frac{10^2}{81^n}\,^{2n}C_2-\frac{10^3}{81^n}\,^{2n}C_3+.....+\frac{10^{2n}}{81^n}is

A

2

B

0

C

12\frac{1}{2}

D

1

Answer

1

Explanation

Solution

The given expression =181n[2nC02nC1101+2nC2102= \frac{1}{81^{n}}[^{2n}C_{0}-\,^{2n}C_{1}\,10^{1}+\,^{2n}C_{2}\,10^{2} 2nC3103+....+2nC2n102n]-\,^{2n}C_{3}\,10^{3}+.... +\,^{2n}C_{2n}\,10^{2n}] =181n[110]2n=(9)2n81n=81n81n=1= \frac{1}{81^{n}}\left[1-10\right]^{2n}= \frac{\left(-9\right)^{2n}}{81^{n}} = \frac{81^{n}}{81^{n}}=1