If $n$ is a positive integer and ${C_k} = {\;^n}{C_k},$ then $\mathop \sum \limits_{k = 1}^n {k^3}{\left( {\frac{{{C_k}}}{{{C_{k - 1}}}}} \right)^2}$ equals

$\frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{{12}}$

$\frac{{n\left( {n + 1} \right){{\left( {n + 2} \right)}^2}}}{{12}}$

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Hard

1 of 1

Analyzing +4 | -1

n

is a positive integer and

{C_k} = {\;^n}{C_k},

then

\sum \limits_{k = 1}^n {k^3}{\left( {\frac{{{C_k}}}{{{C_{k - 1}}}}} \right)^2}

equals

View answer and explanation

Correct option: B

\frac{{n{{\left( {n + 1} \right)}^2}\left( {n + 2} \right)}}{{12}}

Option B (

\frac{{n{{\left( {n + 1} \right)}^2}\left( {n + 2} \right)}}}{{12}}

) is correct because, after simplifying the expression and using the formulas for the sum of first n integers, squares, and cubes, we arrive at the given result.

More Questions from BINOMIAL THEOREM

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\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

equals If

a

and

d

are two complex numbers, then the sum to

\left( {n + 1} \right)

terms of the following series

a{C_0} - \left( {a + d} \right){C_1} + \left( {a + 2d} \right){C_2} - ... + ...

is The number of different terms in the expansion of

(1-x)^{201}\left(1+x+x^2\right)^{200}

is The sum of the coefficient of the binomial expansion of

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

is equal to 6561 . The constant term in the expansion is The coefficient of

{x^5}

in the expansion of

\frac{{1 + {x^2}}}{{1 + x}},\left| x \right| < 1,

is If

\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

, then

n

equals