MereTeacher

JEE • NEET • FOUNDATION

JEE Previous Year Question

$The coefficient of x^{3} in the expansion of (1 - x)^{3/2}, (| x | < 1) is$

Topic: JEE PYQ

Answer: B. Full step-by-step solution below.

A $- \frac{3}{16}$
B $\frac{1}{16}$
C $\frac{1}{8}$
D $\frac{3}{16}$

Correct answer: B

Solution

(b) The expansion of (1 - x)^{3/2} is \Rightarrow (1 - x)^{3/2} = 1 + \frac{3}{2} (- x) + \frac{( \frac{3}{2} ) ( \frac{3}{2} - 1 ) ( x ) ^{2}}{2 !} = 1 - \frac{3}{2} x + \frac{3}{8} x^{2} + \frac{1}{16} x^{3} + \dots ∴ Coefficient of x^{3} = \frac{1}{16}

More Binomial Theorem PYQs

$If A denotes the sum of all the coefficients in the expansion of (1 - 3 x + 10 x^{2})^{n} and B denotes the sum of all the coe\dots$
$If the ratio of the 7th term from the beginning to the 7 th term from the end in the expansion of\dots$
$The remainder when ((64)^{(64)})^{(64)} is divided by 7 is equal to$
$Let m = (9 n^{2} + 54 n + 80) (9 n^{2} + 45 n + 54) (9 n^{2} + 36 n + 35) . The greatest positive integer which divides m, for all positi\dots$
$The term independent of x in the expansion of\dots$