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

To prove that

\[ \sum_{i=1}^{n} (a_i - a)^2 + \sum_{i=1}^{n} a_i b_i = n ab, \]

given that the arithmetic mean of \(a_1, a_2, \ldots, a_n\) is \(a\) and the arithmetic mean of \(b_1, b_2, \ldots, b_n\) is \(b\), and \(a_i + b_i = 1\) for \(i = 1, 2, \ldots, n\), we will follow these steps:

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