What is the value of n?

Question

In this binomial expression $(1+k/6)^n$ what is the value of n given the fact that the coefficient of the 3rd term is twice the coefficient of the 4th term?

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Answer 1 · 2017-03-23T17:02:12

$n=11.$

In the expansion of $(1+x)^n,$ the General $(r+1)^(th)$ Term,

$T_(r+1)=""_nC_rx^r, r=0,1,2,...,n.$

With $x=k/6,$ we have, $T_(r+1)=""_nC_r(k/6)^r, r=0,1,2,...,n.$

$:. r=2 rArr T_3=""_nC_2(k/6)^2=(""_nC_2)/36*k^2," and, similarly,"$

$r=3 rArr T_4=(""_nC_3)/216*k^3.$

Now, by what is given, $(""_nC_2)/36=2{(""_nC_3)/216}.$

$rArr (""_nC_2)/(""_nC_3)=1/3.$

$rArr 3{(n(n-1))/((1)(2))}={(n(n-1)(n-2))/((1)(2)(3))}.$

$rArr 9n(n-1)-n(n-1)(n-2)=0.$

$rArr n(n-1)(9-n+2)=0, or, n(n-1)(11-n)=0.$

$:. n=0, n=1, or, n=11.$

For obvious reason, $n=0, n=1$ are not acceptable.

$:. n=11.$

In the event, the Co-eff. of $T_3$ is $55/36=2(55/72)$

$=2"{the Co-eff. of "$ T_4#}.

Enjoy Maths.!