Find the 4th term in the expansion of (1-x){}³/2}

Correct Answer is : 116x31/16x³161x3

Correct Answer is : 116x31/16x³161x3

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PUNE MCA Exam 2014 Question Paper with answers for online practice

Question : 3 of 58

Marks: +1, -0

{}^{3/2}

Solution:

Concept:

Binomial expansion for fractional exponent:

T_{r+1} = \binom{n}{r} u^r

with

n=\frac{3}{2}

u=-x

Explanation:

The

4

th term corresponds to

r=3

(since

r+1=4

Compute

\binom{3/2}{3} = \frac{(3/2)(1/2)(-1/2)}{3!} = \frac{-3/8}{6} = -\frac{1}{16}

Then

T_4 = \binom{3/2}{3} (-x)^3 = \left(-\frac{1}{16}\right)(-x^3) = \frac{1}{16}x^3

Answer:

Option B:

\frac{1}{16}x^{3}

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