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NCERT Class XI Mathematics - Binomial Theorem - Solutions

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Question : 2 of 36
Marks: +1, -0
(2x−x2)5\left(\frac{2}{x}-\frac{x}{2}\right)^5
Solution:  
We have, (2x−x2)5\left(\frac{2}{x}-\frac{x}{2}\right)^5 = [2x+(−x2)]5\left[\frac{2}{x}+\left(-\frac{x}{2}\right)\right]^5
=  5C0[2x]5\,{}^{5}C_{0}\left[\frac{2}{x}\right]^{5} +  5C1[2x]4[−x2]\,{}^{5}C_{1}\left[\frac{2}{x}\right]^{4}\left[-\frac{x}{2}\right] +  5C2[2x]3[−x2]2\,{}^{5}C_{2}\left[\frac{2}{x}\right]^{3}\left[-\frac{x}{2}\right]^{2} +  5C3[2x]2[−x2]3\,{}^{5}C_{3}\left[\frac{2}{x}\right]^{2}\left[-\frac{x}{2}\right]^{3} +  5C4[2x][−x2]4\,{}^{5}C_{4}\left[\frac{2}{x}\right]\left[-\frac{x}{2}\right]^{4} +  5C5[−x2]5\,{}^{5}C_{5}\left[-\frac{x}{2}\right]^{5}
= 1 [32x5]+5(8x3)(−1)\left[\frac{32}{x^5}\right] + 5\left(\frac{8}{x^3}\right)(-1) + 10 [2x]\left[\frac{2}{x}\right] + 10 [−x2]+5[x38]+[−x532]\left[-\frac{x}{2}\right] + 5\left[\frac{x^3}{8}\right] + \left[-\frac{x^5}{32}\right]
= 31x5−40x3+20x\frac{31}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + 5x38−x532\frac{5x^3}{8} - \frac{x^5}{32}
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