Question

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`?

Answer 1

`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`