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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
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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