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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

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                                                                                         

Grade:

1 Answers

Mrinal Shingavi
18 Points
9 years ago

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

b(n)=nC1+(nC2)a+(nC3)a2........(nCn)an-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={41/401-1}

We get {b(2006)-b(2005)} =211

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