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NCERT Class XI Mathematics - Binomial Theorem - Solutions

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Question : 35 of 36
Marks: +1, -0
Expand using Binomial Theorem (1+x2−2x)4\left(1+\frac{x}{2}-\frac{2}{x}\right)^4 , x ≠ 0
Solution:  
Let x2−2x\frac{x}{2}-\frac{2}{x} = y
∴ (1+x2−2x)4\left(1+\frac{x}{2}-\frac{2}{x}\right)^4 = (1+y)4(1+y)^4
= (40)+(41)y+(43)y3+(44)y4\binom{4}{0}+\binom{4}{1}y+\binom{4}{3}y^3+\binom{4}{4}y^4
= (40)+(41)[x2−2x]+(42)[x2−2x]2\binom{4}{0}+\binom{4}{1}\left[\frac{x}{2}-\frac{2}{x}\right]+\binom{4}{2}\left[\frac{x}{2}-\frac{2}{x}\right]^2 + (43)[x2−2x]3+(44)[x2−2x]4\binom{4}{3}\left[\frac{x}{2}-\frac{2}{x}\right]^3+\binom{4}{4}\left[\frac{x}{2}-\frac{2}{x}\right]^4
= 1 + 4 [x2−2x]\left[\frac{x}{2}-\frac{2}{x}\right] +
6[(20)(x2)2+(21)(x2)(−2x)+(22)(−2x)2]6\left[\binom{2}{0}\left(\frac{x}{2}\right)^2+\binom{2}{1}\left(\frac{x}{2}\right)\left(-\frac{2}{x}\right)+\binom{2}{2}\left(-\frac{2}{x}\right)^2\right]
+
4[(30)(x2)3+(31)(x2)2(−2x)+(32)(x2)(−2x)2+(33)(−2x)3]4\left[\binom{3}{0}\left(\frac{x}{2}\right)^3+\binom{3}{1}\left(\frac{x}{2}\right)^2\left(-\frac{2}{x}\right)+\binom{3}{2}\left(\frac{x}{2}\right)\left(-\frac{2}{x}\right)^2+\binom{3}{3}\left(-\frac{2}{x}\right)^3\right]
+
[(40)(x2)4+(41)(x2)3(−2x)+(42)(x2)2(−2x)2+(43)(x2)(−2x)3+(44)(−2x)4]\left[\binom{4}{0}\left(\frac{x}{2}\right)^4+\binom{4}{1}\left(\frac{x}{2}\right)^3\left(-\frac{2}{x}\right)+\binom{4}{2}\left(\frac{x}{2}\right)^2\left(-\frac{2}{x}\right)^2+\binom{4}{3}\left(\frac{x}{2}\right)\left(-\frac{2}{x}\right)^3+\binom{4}{4}\left(-\frac{2}{x}\right)^4\right]
= 1 + 2x - 8x+6[x24−2+4x2]\frac{8}{x}+6\left[\frac{x^2}{4}-2+\frac{4}{x^2}\right] + 4[x38−3x2+6x−8x3]4\left[\frac{x^3}{8}-\frac{3x}{2}+\frac{6}{x}-\frac{8}{x^3}\right] + [x416−x2+6−16x2+16x4]\left[\frac{x^4}{16}-x^2+6-\frac{16}{x^2}+\frac{16}{x^4}\right]
= 1 + 2x - 8x+32x2\frac{8}{x}+\frac{3}{2}x^2 - 12 + 24x2+x32\frac{24}{x^2}+\frac{x^3}{2} - 6x + 34x−32x3+x416−x2\frac{34}{x}-\frac{32}{x^3}+\frac{x^4}{16}-x^2 + 6 - 16x2+16x4\frac{16}{x^2}+\frac{16}{x^4}
= 16x4−32x3+8x2+16x\frac{16}{x^4}-\frac{32}{x^3}+\frac{8}{x^2}+\frac{16}{x} - 5 - 4x + x22+x32+x416\frac{x^2}{2}+\frac{x^3}{2}+\frac{x^4}{16}
= 16x+8x2−32x3+16x4\frac{16}{x}+\frac{8}{x^2}-\frac{32}{x^3}+\frac{16}{x^4} - 4x + x22+x32+x416−5\frac{x^2}{2}+\frac{x^3}{2}+\frac{x^4}{16}-5
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