JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Easy

1 of 1

Analyzing +4 | -1

The value of

\frac{1}{{{{81}^n}}} - \frac{{10}}{{{{81}^n}}}{\;^{2n}}{C_1}\;\frac{{{{10}^2}}}{{{{81}^n}}}{\;^{2n}}{C_2} - \frac{{{{10}^3}}}{{{{81}^n}}}2{\;^{2n}}{C_3} + ... + \frac{{{{10}^{2n}}}}{{{{81}^n}}}\;\mathrm{is}

View answer and explanation

Correct option: D

The correct answer is D (1) because the expression can be recognized as a binomial expansion, which simplifies to 1.

More Questions from BINOMIAL THEOREM

Practice All Chapter MCQs →

\left({ }^{10} C_0\right)^2-\left({ }^{10} C_1\right)^2+\ldots-\left({ }^{10} C_9\right)^2+\left({ }^{10} C_{10}\right)^2

a

d

\left( {n + 1} \right)

a{C_0} - \left( {a + d} \right){C_1} + \left( {a + 2d} \right){C_2} - ... + ...

(1-x)^{201}\left(1+x+x^2\right)^{200}

\left(\frac{1}{x}+2 x\right)^n

{x^5}

\frac{{1 + {x^2}}}{{1 + x}},\left| x \right| < 1,

\left(\begin{array}{c}2 n+1 \\ 0\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 3\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 6\end{array}\right)+\ldots=170

n