JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Analyzing +4 | -1

The value of

1+\left(\frac{1}{2}+\frac{1}{3}\right) \frac{1}{4}+\left(\frac{1}{4}+\frac{1}{5}\right) \frac{1}{4^2}+\left(\frac{1}{6}+\frac{1}{7}\right) \frac{1}{4^3}+\ldots

View answer and explanation

Correct option: A

\log _e \sqrt{12}

Option A (

\log _e \sqrt{{12}}

) is correct because the given infinite series can be simplified and expressed in terms of logarithms, ultimately resulting in

\log _e \sqrt{{12}}

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