Expand the following-x-3-1-x5

Using binomial theorem for the expansion of ( x/3+1/x )^5, we have ( x/3+1/x )^5 = 5_C_0 ( x/3 )^5+ 5_C_1 ( x/3 )^4 ( 1/x ) + 5_C_2 ( x/3 )^3 ( 1/x )^2+ ...

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Byju's Answer

Standard XI

Mathematics

Multinomial Expansion

Expand the fo...

Question

Expand the following:
${(\frac{x}{3} + \frac{1}{x})}^{5}$

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Solution

Using binomial theorem for the expansion of ${(\frac{x}{3} + \frac{1}{x})}^{5}$ , we have

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

$= \frac{x^{5}}{243} + 5. \frac{x^{4}}{81} . \frac{1}{x} + 10. \frac{x^{3}}{27} . \frac{1}{x^{2}} + 10. \frac{x^{2}}{9} . \frac{1}{x^{3}} + 5. \frac{x}{3} . \frac{1}{x^{4}} + \frac{1}{x^{5}}$

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

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101

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Question 64

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Expand the following: (x3+1x)5

Using binomial theorem for the expansion of (x3+1x)5, we have (x3+1x)5=5C0(x3)5+5C1(x3)4(1x)+5C2(x3)3(1x)2+ 5C3(x3)2(1x)3+5C4(x3)(1x)4+5C5(1x)5 =x5243+5.x481.1x+10.x327.1x2+10.x29.1x3+5.x3.1x4+1x5 =x5243+581x3+1027x+109x+53x3+1x5

Expand the following:
${(\frac{x}{3} + \frac{1}{x})}^{5}$