JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Easy

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Analyzing +4 | -1

Sum of the infinite series

1 + \frac{1}{3}.\frac{1}{2} + \frac{2}{3}.\frac{5}{6}.\frac{1}{{{2^2}}} + \frac{2}{3}.\frac{5}{6}.\frac{8}{9}.\frac{1}{{{2^3}}} + \ldots \infty \;\mathrm{is}

View answer and explanation

Correct option: B

{4^{1/3}}

Option B (

{4^{1/3}}

) is correct because by comparing the given series with the binomial expansion of

(1+x)^n

, we can determine the values of

n

and

x

, and then find the sum.

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