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+1 vote
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in Binomial Theorem by (22.8k points)
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The sum of coefficients of the expansion \((\frac{1}{x}+2x)^n\) is 6561. The coefficient of term independent of x is

(a) 16. 8C

(b) 8C4 

(c) 8C5 

(d) None of these

1 Answer

+2 votes
by (23.7k points)
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Best answer

Answer : (a) 16. 8C4   

Sum of the coefficients of the expansion \((\frac{1}{x} +2x)^n\)  = 6561

Putting x = 1, (1 + 2)n = 6561 ⇒ 3n = 38 

⇒ n = 8

 \(\therefore\) Tr + 1 in the expansion of \((\frac{1}{x} +2x)^8\) 

= 8Cr \((\frac{1}{x})^{8-r}\) (2x)r = 8Cr 2x2r-8 

Since this term is independent of x, 2r – 8 = 0 ⇒ r = 4.

∴ Reqd. term = 8C4 . 24 

= 16 . 8C4.

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