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If the last term in the expansion of $\bigg(2^{\large\frac{1}{3}}-\frac{1}{\sqrt{ 2}}\bigg)^n $ is $\bigg(\large\frac{1}{3^{\large\frac{5}{3}}}\bigg)^{log_38}$, then the $5^{th} $ term from the beginning is ?

(A) 840

(B) 210

(C) 320

(D) 420

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  • $log_a a=1$
  • $log a^n=n.log a$
  • $x^{log_x y}=y$
  • General term in the expansion of $(x+y)^n=T_{r+1}=^nCr.x^{n-r}.y^r$
The last terrm in the expansion of $\bigg(2^{1/3}-\large\frac{1}{\sqrt 2}$$\bigg)^n$ is
$(-1)^n.^nC_n.\big(2^{ 1/3}\big)^{n-n}.\big(\large\frac{1}{\sqrt 2}\big)^n$
$\bigg(\large\frac{1}{3^{5/3}}$$\bigg)^{log_3 8}=3^{\large\frac{-5log_3 8}{3}}=3^{\large\frac{-5.3.\: log_3 2}{3}}$
$=3^{log_3 2^{-5}}$ $=2^{-5}$
Given: $\large\frac{(-1)^n}{2^{n/2}}$$=2^{-5}$
$5^{th} $ term= $^{10}C_4.2^{6/3}.\big(-\large\frac{1}{\sqrt 2}\big)^4$


answered Oct 20, 2013 by rvidyagovindarajan_1

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