Question

Let $a_i+b_i=1(i=1,2,..n)$ and $a=\frac{1}{n}(a_1+a_2+....+a_n)$ , $b=\frac{1}{n}(b_1+b_2+....+b_n)$ then $a_1{b}_1+a_2{b}_2+..+.a_n{b}_n$is equal to

• A. $nab$
• B. $nab–{(a_1-a)}^2–{(a_2–a)}^2-\dots {(a_n–a)}^2$
• C. ${(a_1-a)}^2–{(a_2–a)}^2-\dots {(a_n–a)}^2-nab$
• D. None of these

Answer 1

The correct option is B

$nab–{(a_1-a)}^2–{(a_2–a)}^2-\dots {(a_n–a)}^2$

Finding the value of $a_1{b}_1+a_2{b}_2+..+.a_n{b}_n$

We can write,

$$\begin{gathered} a_1{b}_1+a_2{b}_2+..+.a_n{b}_n=\sum _{i=1}^n{a}_i{b}_i \\ =\sum _{i=1}^n{a}_i(1-a_i)\left[\because a_i+b_i=1\Rightarrow b_i=a_i-1\right] \\ =\sum _{i=1}^n{a}_i–\sum _{i=1}^n{a_i}^2 \\ =an–\sum _{i=1}^n{\left(\left(a_i-a\right)+a\right)}^2\left[\because a=\frac{1}{n}(a_1+a_2+....+a_n)\Rightarrow an=\sum _{i=1}^n{a}_i\&\text{adding and subtracting}a\right] \\ =an–\sum _{i=1}^n[{(a_i–a)}^2+a^2+2a(a_i-a\left)\right]\left[\because {\left(a+b\right)}^2=a^2+b^2+2ab\right] \\ =an–\sum _{i=1}^n{(a_i–a)}^2+\sum _{i=1}^n{a}^2–2a\sum _{i=1}^n(a_i-a) \\ =an–\sum _{i=1}^n{(a_i–a)}^2+n{a}^2–2a\sum _{i=1}^n(a_i-a)\left[\because a=\frac{1}{n}(a_1+a_2+....+a_n)\Rightarrow an=\sum _{i=1}^n{a}_i\right] \\ =an(1-a)–\sum _{i=1}^n{(a_i–a)}^2–2a\sum _{i=1}^n(a_i-a) \\ =nab–\sum _{i=1}^n{(a_i–a)}^2–2a(na-na)\left[\because b_i=a_i-1\right] \\ =nab–\sum _{i=1}^n{(a_i–a)}^2 \\ =nab-{(a_1–a)}^2–{(a_2–a)}^2–\dots (a_n–a) \end{gathered}$$Hence, option B is correct.