If ${\left( {1 + x} \right)^n} = {C_0} + {C_1}x + {C_2}{x^2} + ... + {C_n}{x^n}$, then $\begin{array}{*{20}{c}} {\sum \;}\\ {0 \le r < s \le n} \end{array}\;\;\sum \left( {r + s} \right){C_r}{C_s}\;$ is equal to

$2n\left[ {{2^{2n - 1}}{ - ^{2n - 1}}{C_{n - 1}}} \right]$

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Easy

1 of 1

Analyzing +4 | -1

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

, then

\begin{array}{cccccccccccccccccccc} {\sum \;}\\ {0 \le r < s \le n} \end{array}\;\;\sum \left( {r + s} \right){C_r}{C_s}\;

is equal to

View answer and explanation

Correct option: A

n\left[ {{2^{2n - 1}}{ - ^{2n - 1}}{C_{n - 1}}} \right]

Option A (

n[2^{2n-1} - ^{2n-1}C_{n-1}]

) is correct because it is derived by manipulating the sum of products of binomial coefficients.

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