If `b_i=1-a_i ,n a=sum_(i=1)^n a_i ,n b=sum_(i=1)^n b_i ,t h e nsum_(i=1)^n a_i ,b_i+sum_(i=1)^n(a_i-a)^2=` `a b` b. ` n a b` c. `(n+1)a b` d. `n a b`

Options

A. ab
B. `-nab`
C. `(n+1)ab`
D. nab

Answer

D

Explanation

`Sigmaa_(i)b_(i)=Sigmaa_(i)(1-a_(i))`
`=na-Sigmaa_(i)^(2)`
`=na-Sigma(a_(i)-a+a)^(2)`
`=na-Sigma[(a_(i)-a)^(2)+a^(2)+2a(a_(i)-a)]`
`=na-Sigma(a_(i)-a)^(2)-Sigmaa^(2)-2aSigma(a_(i)-a)`
`rArrSigmaa_(i)b_(i)+Sigma(a_(i)-a)^(2)=na-na^(2)-2a(na-na)`
`=na(1-a)=nab` `[{:(becauseSigmab_(i)=Sigma1-Sigmaa_(i)),(thereforenb=n-na),(or a+b=1):}]`