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Algebraic Identities

If b i =1- ...

Question

If $b_{i} = 1 - a_{i}, n a = \sum_{i = 1}^{n} a_{i}, n b = \sum_{i = 1}^{n} b_{i}$ , then $\sum_{i = 1}^{n} {a_{i} b}_{i} + \sum_{i = 1}^{n} {(a_{i} - a)}^{2} =$

A

a b

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B

- n a b

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C

n a b

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D

(n + 1) a b

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Solution

The correct option is C $n a b$
$a_{1} + b_{1} = 1 n a = a_{1} + a_{2} + a_{3} + . . . + a_{n} (1) a_{2} + b_{2} = 1 n b = b_{1} + b_{2} + b_{3} + . . . . + b_{n} (2) a_{3} + b_{3} = 1 - - - - - a_{n} + b_{n} = 1 - ----------- - (a_{1} + a_{2} + a_{3} + . . . + a_{n}) + (b_{1} + b_{2} + b_{3} + . . . . + b_{n}) = n (3)$
From $(1), (2), (3)$
$n a + n b = n a + b = 1 (4)$
$\sum_{i = 1}^{n} a_{i} b_{i} + \sum_{i = 1}^{n} {(a_{i} - a)}^{2} \Rightarrow \sum_{i = 1}^{n} a_{i} (1 - a_{i}) + \sum_{i = 1}^{n} {a_{i}}^{2} + \sum_{i = 1}^{n} a^{2} - 2 \sum_{i = 1}^{n} a_{i} a \Rightarrow \sum_{i = 1}^{n} a_{i} - \sum_{i = 1}^{n} {a_{i}}^{2} + \sum_{i = 1}^{n} {a_{i}}^{2} + a^{2} \sum_{i = 1}^{n} (1) - 2 a \sum_{i = 1}^{n} a_{i} = (1 - 2 a) \sum_{i = 1}^{n} a_{i} + a^{2} \sum_{i = 1}^{n} (1) = (1 - 2 a) (n a) + a^{2} (n) = n a - 2 a^{2} n + a^{2} n = n a - a^{2} n = n a (1 - a) = n a b f r o m (4)$

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