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NCERT Class XI Mathematics - Binomial Theorem - Solutions

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Question : 4 of 36
Marks: +1, -0
(x3+1x)5\left(\frac{x}{3}+\frac{1}{x}\right)^5
Solution:  
We have, (x3+1x)5\left(\frac{x}{3}+\frac{1}{x}\right)^5
=  5C0(x3)5\,{}^{5}C_{0}\left(\frac{x}{3}\right)^5 +  5X1(x3)4(1x)\,{}^{5}X_{1}\left(\frac{x}{3}\right)^4\left(\frac{1}{x}\right) +  5C2(x3)3(1x)2\,{}^{5}C_{2}\left(\frac{x}{3}\right)^3\left(\frac{1}{x}\right)^2 +  5C3(x3)2(1x)3\,{}^{5}C_{3}\left(\frac{x}{3}\right)^2\left(\frac{1}{x}\right)^3 +  5C4(x3)(1x)4\,{}^{5}C_{4}\left(\frac{x}{3}\right)\left(\frac{1}{x}\right)^4 +  5C5(1x)5\,{}^{5}C_{5}\left(\frac{1}{x}\right)^5
= x5243+5x381+10x7\frac{x^5}{243}+\frac{5x^3}{81}+\frac{10x}{7} + 109x+53x3+1x5\frac{10}{9x}+\frac{5}{3x^3}+\frac{1}{x^5}
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