JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Analyzing +4 | -1

(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots+C_n x^n

, then

{\sum\limits_{0 \le i} {\sum\limits_{ < j \le n} {({C_i} - {C_j})} } ^2}

View answer and explanation

Correct option: A

(n+1){ }^{2 n} C_n-2^{2 n}

Option A is correct because the given summation simplifies to

(n+1){ }^{2 n} C_n-2^{2 n}

using properties of binomial coefficients.

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