JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Application +4 | -1

(1+{x})^{{n}}={c}_0+{c}_1+{c}_2 {X}^2 \ldots {c}_{{n}} {X}^{{n}}

then the values of

\sum\limits_{{k=0}}^n(k+1)^2 c_k

View answer and explanation

Correct option: B

2^{{n}-2}\left(n^2+5 n+4\right)

Option B (

2^{{n}-2}\left(n^2+5 n+4\right)

) is correct because it is the result of applying differentiation and multiplication by x to the binomial expansion and then substituting x=1.

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

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