Question
${(1 - 2x)^{3/2}}$ के प्रसार में $ 4$ वाँ पद होगा
$ = 1 + \frac{3}{2}( - 2x) + \frac{3}{2}.\frac{1}{2}.\frac{1}{2}{( - 2x)^2} + \frac{3}{2}.\frac{1}{2}\left( { - \frac{1}{2}} \right)\frac{1}{6}{( - 2x)^3} + .....$
अत: $4$ वाँ पद $\frac{{{x^3}}}{2}$ है
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