JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Easy

1 of 1

Application +4 | -1

{\left( {1 + x} \right)^n} = {\;^n}{C_0} + {\;^n}{C_1}x + {\;^n}{C_2}{x^2} + ... + {\;^n}{C_n}{x^n},

then

\frac{{{\;^n}{C_1}}}{{{\;^n}{C_0}}} + \frac{{2{\;^n}{C_2}}}{{{\;^n}{C_1}}} + \frac{{\;{3^n}{C_3}}}{{{\;^n}{C_2}}} + ... + \frac{{n{\;^n}{C_n}}}{{{\;^n}{C_{n - 1}}}}

is equal to

View answer and explanation

Correct option: C

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

The correct answer is

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

because substituting values for n and checking the options reveals that only option C satisfies the condition.

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