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.!