If the third term in the binomial expansion of ${\left( {1 + x} \right)^m}$ is $ - \frac{1}{8}{x^2},$ then the rational value of $m$, is

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Easy

1 of 1

Application +4 | -1

If the third term in the binomial expansion of

{\left( {1 + x} \right)^m}

- \frac{1}{8}{x^2},

then the rational value of

m

, is

View answer and explanation

Correct option: B

\frac{1}{2}

Option B (1/2) is correct because by setting the third term of the binomial expansion equal to -1/8 x², we can solve for m.

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