The fourth term in the expansion of ( 1 − 2 x ) 3 2 will be Options: A . &nbsp - 3 4 x 4 B . &nbsp x 3 2 C . &nbsp - x 3 2 D . &nbsp 3 4 x 4

Answer: Option B : B Expansion of ( 1 − 2 x ) 3 2 = 1 + 3 2 ( − 2 x ) + 3 2 . 1 2 . 1 2 ( − 2 x ) 2 + 3 2 . 1 2 ( − 1 2 ) 1 6 ( − 2 x ) 3 + . . . . . . . . . Hence 4 t h term is x 3 2

The fourth term in the expansion of (1−2x)32 will be

Discussion Forum : Binomial Theorem

Question -

The fourth term in the expansion of $(1 - 2 x)^{\frac{3}{2}}$ will be

Options:

A . -

\frac{3}{4}

x^{4}

B .

\frac{x^{3}}{2}

C . -

\frac{x^{3}}{2}

D .

\frac{3}{4}

x^{4}

Answer: Option B
:
B

Expansion of $(1 - 2 x)^{\frac{3}{2}}$

= $1 + \frac{3}{2} (- 2 x) + \frac{3}{2} . \frac{1}{2} . \frac{1}{2} (- 2 x)^{2} + \frac{3}{2} . \frac{1}{2} (- \frac{1}{2}) \frac{1}{6} (- 2 x)^{3} + . . . . . . . . .$

Hence $4^{t h}$ term is $\frac{x^{3}}{2}$

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Discussion Forum : Binomial Theorem

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