( x 3 + 1 x ) 5 =

Question

${(\frac{x}{3} + \frac{1}{x})}^{5} =$

Smart Achievers · Accepted Answer

(1)

$We have, {(\frac{x}{3} + \frac{1}{x})}^{5}$

$= {}^{5}C_{0} {(\frac{x}{3})}^{5} + {}^{5}C_{1} {(\frac{x}{3})}^{4} (\frac{1}{x}) + {}^{5}C_{2} {(\frac{x}{3})}^{3} {(\frac{1}{x})}^{2}$ $+ {}^{5}C_{3} {(\frac{x}{3})}^{2} {(\frac{1}{x})}^{3} + {}^{5}C_{4} (\frac{x}{3}) {(\frac{1}{x})}^{4} + {}^{5}C_{5} {(\frac{1}{x})}^{5}$

$= \frac{x^{5}}{243} + \frac{5 x^{3}}{81} + \frac{10 x}{27} + \frac{10}{9 x} + \frac{5}{3 x^{3}} + \frac{1}{x^{5}}$

Solution

Q.
${(\frac{x}{3} + \frac{1}{x})}^{5} =$

1 $\frac{x^{5}}{243} + \frac{5 x^{3}}{81} + \frac{10 x}{27} + \frac{10}{9 x} + \frac{5}{3 x^{3}} + \frac{1}{x^{5}}$

2 $\frac{x^{5}}{420} + \frac{5 x^{4}}{2} + \frac{3 x^{3}}{81} + \frac{10 x}{27} + \frac{5}{x} + \frac{2}{x^{4}}$

3 $\frac{2 x^{5}}{243} + \frac{4 x^{4}}{3} + \frac{3 x^{3}}{81} + \frac{10 x}{9} + \frac{10}{9 x} + \frac{5}{3 x^{3}}$

4 None of these

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Solution

Q. (x3+1x)5=

1 x5243+5x381+10x27+109x+53x3+1x5

2 x5420+5x42+3x381+10x27+5x+2x4

3 2x5243+4x43+3x381+10x9+109x+53x3