JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Medium

1 of 1

Analyzing +4 | -1

(1+x)\left(1+x+x^2\right)\left(1+x+x^2+x^3\right) \ldots\left(1+x+x^2+\ldots+x^{100}\right)

when written in the ascending power of

x

, then the highest exponent of

x

\lambda(1010)

, where

\lambda

equals___.

View answer and explanation

Correct option: A

Option A (5) is correct because the highest exponent is the sum of the exponents from 1 to 100, which is 5050, and 5050 divided by 1010 is 5.

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