If the arithmetic mean of `a_(1),a_(2),a_(3),"........"a_(n)` is a and `b_(1),b_(2),b_(3),"........"b_(n)` have the arithmetic mean b and `a_(i)+b_(i)=1` for `i=1,2,3,"……."n,` prove that `sum_(i=1)^(n)(a_(i)-a)^(2)+sum_(i=1)^(n)a_(i)b_(i)=nab`.

Explanation

`suma_(i)b_(i)=suma_(i)(1-a_(i))=na-suma_(i)^(2)`
`=na-sum(a_(i)-a+a)^(2)`
`=na-sum[(a_(i)-a)^(2)+a^(2)+2a(a_(i)-a)]`
`=na-sum(a_(i)-a)^(2)+suma^(2)+2asum(a_(i)-a)`
`:.suma_(i)b_(i)+sum(a_(i)-a)^(2)=na-na^(2)-2a(na-na)`
`=na(1-a)=nab [{:(sumb_(1)=sum1-suma_(i)),(therefore nb=n-na),(" or "a+b=1):}]`.