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?

4155 views around the world

You can reuse this answer
Creative Commons License

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.!