JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Application +4 | -1

(1+\mathrm{x})^n=\mathrm{C}_0+\mathrm{C}_1 \mathrm{x}+\mathrm{C}_2 \mathrm{x}^2+\ldots+\mathrm{C}_n \mathrm{x}^{\mathrm{n}}

, then

\mathrm{C}_0 \mathrm{C}_1+\mathrm{C}_1 \mathrm{C}_2+\mathrm{C}_2 \mathrm{C}_3+\ldots .+\mathrm{C}_{n-1} \mathrm{C}_n

is equal to-

View answer and explanation

Correct option: C

\frac{2 n !}{(n-1) !(n+1) !}

Option C (

\frac{{2n!}}{{(n-1)!(n+1)!}}

) is correct because the given expression can be interpreted as the coefficient of

x^{n-1}

in the expansion of

(1+x)^n(1+x)^n

, which simplifies to

{}^{2n}C_{n-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