JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Analyzing +4 | -1

The value of

\sum\limits_{1 \le i < } {\sum\limits_{j \le n - 1} {{{(i.j)}^n}{C_i}{.^n}{C_j}} }

is equal to

View answer and explanation

Correct option: A

\frac{n^2}{2}\left(2^{2 n-2}-{ }^{2 n-2} C_{n-1}\right)

Option A is correct because the summation can be simplified using properties of binomial coefficients and results in

\frac{n^2}{2}\left(2^{2 n-2}-{ }^{2 n-2} C_{n-1}\right)

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