JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Easy

1 of 1

Application +4 | -1

Sum of the series

2 \mathrm{C}_0+\frac{\mathrm{C}_1}{2} 2^2+\frac{\mathrm{C}_2}{3} 2^3+\frac{\mathrm{C}_3}{4} 2^4+\ldots \ldots \cdot \frac{\mathrm{C}_{\mathrm{n}}}{\mathrm{n}+1} 2^{\mathrm{n}+1}

equals-

View answer and explanation

Correct option: B

\frac{3^{\mathrm{n}+1}-1}{\mathrm{n}+1}

The correct answer is B (

\frac{{3^{n+1}-1}}{{n+1}}

) because the given series can be expressed as a summation involving binomial coefficients, which simplifies to the given expression.

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