If `(1+x)^n=sum_(r=0)^n a_r x^r and b_r=1+(a_r)/(a_(r-1)) and prod_(r=1)^n b_r=((101)^(100))/(100!),` then n equals to: 99 (b) 100 (c) 101 (d) None of these

Explanation

Here, `a_(r) = ""^(n)C_(r)`
`therefore b_(r) = 1 +(a_(r))/(a_(1 -1)) = 1 + (""^(n)C_(r))/(""^(n)C_(r-1))`
`= 1 + (n-r+1)/r = (n+1)/r`
`rArr prod_(r=1)^(n) b_(r) = prod_(r=1)^(n) ((n+1))/r `
`= ((n+1))/1 cdot ((n+1))/2 cdot ((n+1))/3 ... ((n+1))/n= ((n+1)^(n))/(n!)`
`=((101)^(100))/(100!) ` [ giben]
`therefore n= 100 rArr n/20 = 5`