Let a={ 4^[1/401]-1} and for each n greater than equal to 2 let [ b]n =nC1+[nC2]a+[nC3]a^2 +----------+[nCn]a^n-1 find the value of{ [b]2006-[b]2005}? reply quickly

first of all convert the series of b(n) into binomial theorem.

b(n)=nC1+(nC2)a+(nC3)a²........(nCn)a^n-1

      =[(1+a)ⁿ - nC0]/a

{b(2006)-b(2005)} =[(1+a)²⁰⁰⁶ -(1+a)²⁰⁰⁵-1]/a

                             ={(1+a)²⁰⁰⁵[(1+a)-1]-1} /a

                            ={a(1+a)²⁰⁰⁵-1}/a

Now putting value of a that is a={4^1/401-1}

We get {b(2006)-b(2005)} =2¹¹