If the ratio of the 7th term from the beginning to the 7 th term from the end in the expansion of ( [3]2+ 1 [3]3)^n is 16, then n=

Question

Accepted Answer

Correct answer: C. (c) Here, the general term in the expansion of ( [3]2+ 1 [3]3)^n is T_r+1=^nC_r (2^1/3 )^n-r ( 13^1/3 )^r Now, the 7 th term from beginning =^nC_62^ n-633^-2 and the 7 th term from end =^nC_6 ( 13^1/3 )^n-6 (2^1/3 )^6 Hence, ratio = ^nC_62^ n-633^-2^nC_63^ 6-n32^2= 16 ⇒ 1 6-n6^3+2= 16 ⇒ 6-n3+2=1 ⇒ 6-n3=-1 ⇒6-n=-3 ⇒n=9 Hence, option (c) is correct.

$If the ratio of the 7th term from the beginning to the 7 th term from the end in the expansion of (32 + \frac{1}{3 3})^{n} is \frac{1}{6}, then n =$

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