JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Hard

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Analyzing +4 | -1

(1+x)^n=\sum\limits_{{r=0}}^n{ }^n C_r x^n

, then

\frac{C_0}{1 \cdot 2} 2^2+\frac{C_1}{2 \cdot 3} 2^3

+\frac{C_2}{3 \cdot 4} 2^4+\ldots+\frac{C_n}{(n+1)(n+2)} 2^{n+2}

is equal to

View answer and explanation

Correct option: C

\frac{3^{n+2}-2 n-5}{(n+1)(n+2)}

Option C (

\frac{{3^{n+2}-2 n-5}}{{(n+1)(n+2)}}

) is correct because it is the sum of the series after applying binomial theorem and simplifying the 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