In the expansion of #(a+x)(1-x/4)^n#, the first three terms are #4+bx+5/4x^2#, where #n# is a positive integer greater than #2#. Find the values of #n#, of #a# and of #b#?

1 Answer
Feb 20, 2018

#n=5#, #a=4# and #b=-4#

Explanation:

Using binomial expansion #(1-x/4)^n#

= #1-n*x/4+(n(n-1))/2(x^2/16)-(n(n-1)(n-2))/6(x^3/64)+......#

and #(a+x)(1-x/4)^n#

= #(a+x)(1-n*x/4+(n(n-1))/2(x^2/16)-(n(n-1)(n-2))/6(x^3/64)+......)# - (listing only first three terms i.e. constant term and containing #x# and#x^2#)

= #a+x-(nax)/4-(nx^2)/4+(n(n-1))/2(x^2/16)+a(n(n-1))/2(x^2/16)+............#

= #a+x(1-(na)/4)+x^2(-n/4+(an(n-1))/32)+.........#

As first three terms are #4+bx+5/4x^2#, we have

#a=4#, #1-(na)/4=b# and #-n/4+(an(n-1))/32=5/4#

Putting #a=4# in the last equation we get

#-n/4+(n(n-1))/8=5/4# or #-2n+n^2-n-10=0#

or #n^2-3n-10=0# i.e. #(n-5)(n+2)=0#

and as #n# is a positive integer, we have #n=5#

and as #1-(na)/4=b#, we have #b=1-5=-4#

Hence, #n=5#, #a=4# and #b=-4#