Question

Prove that\(\sum\limits_{{\bf{i = 1}}}^{\bf{n}} {{{\left( {{{\bf{c}}_{\bf{1}}}{{\bf{x}}_{\bf{i}}}{\bf{ + }}{{\bf{c}}_{\bf{2}}}} \right)}^{\bf{2}}}} {\bf{ = c}}_{\bf{1}}^{\bf{2}}\sum\limits_{{\bf{i = 1}}}^{\bf{n}} {{{\left( {{{\bf{x}}_{\bf{i}}}{\bf{ - \bar x}}} \right)}^{\bf{2}}}} {\bf{ + n}}{\left( {{{\bf{c}}_{\bf{1}}}{\bf{\bar x + }}{{\bf{c}}_{\bf{2}}}} \right)^{\bf{2}}}\).

Short answer

It is proved that, \(\sum\limits_{i = 1}^n {{{\left( {{c_1}{x_i} + {c_2}} \right)}^2}} = c_1^2\sum\limits_{i = 1}^n {{{\left( {{x_i} - \bar x} \right)}^2}} + n{\left( {{c_1}\bar x + {c_2}} \right)^2}\)

Step-by-step solution

State the general expression

To prove:

\(\sum\limits_{i = 1}^n {{{\left( {{c_1}{x_i} + {c_2}} \right)}^2}} = c_1^2\sum\limits_{i = 1}^n {{{\left( {{x_i} - \bar x} \right)}^2}} + n{\left( {{c_1}\bar x + {c_2}} \right)^2}\)

The formula used to expand the expression,

\({\bf{\bar x = }}\frac{{\bf{1}}}{{\bf{n}}}\sum\limits_{{\bf{i = 1}}}^{\bf{n}} {{{\bf{x}}_{\bf{i}}}} \)then\(\sum\limits_{{\bf{i = 1}}}^{\bf{n}} {{{\bf{x}}_{\bf{i}}}} {\bf{ = n\bar x}}\).

Expand the expression \(\sum\limits_{{\bf{i = 1}}}^{\bf{n}} {{{\left( {{{\bf{c}}_{\bf{1}}}{{\bf{x}}_{\bf{i}}}{\bf{ + }}{{\bf{c}}_{\bf{2}}}} \right)}^{\bf{2}}}} \)

Expanding the left side first, we have that, \(\begin{array}{c}\sum\limits_{i = 1}^n {{{\left( {{c_1}{x_i} + {c_2}} \right)}^2}} = \sum\limits_{i = 1}^n {\left( {c_1^2x_i^2 + 2{c_1}{c_2}{x_i} + c_2^2} \right)} \\ = c_1^2\sum\limits_{i = 1}^n {x_i^2} + 2{c_1}{c_2}\sum\limits_{i = 1}^n {{x_i}} + nc_2^2\;\;\;\;\;\;\;\;\;\;\;\left( {\sum\limits_{i = 1}^n {{x_i}} = n\bar x} \right)\\ = c_1^2\sum\limits_{i = 1}^n {x_i^2} + 2n{c_1}{c_2}\bar x + nc_2^2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;...\left( 1 \right)\;\;\;\;\end{array}\)

Expand the expression \({\bf{c}}_{\bf{1}}^{\bf{2}}\sum\limits_{{\bf{i = 1}}}^{\bf{n}} {{{\left( {{{\bf{x}}_{\bf{i}}}{\bf{ - \bar x}}} \right)}^{\bf{2}}}} {\bf{ + n}}{\left( {{{\bf{c}}_{\bf{1}}}{\bf{\bar x + }}{{\bf{c}}_{\bf{2}}}} \right)^{\bf{2}}}\)

`Now, expanding the right side, we have that,

\(\begin{array}{c}c_1^2\sum\limits_{i = 1}^n {{{\left( {{x_i} - \bar x} \right)}^2}} + n{\left( {{c_1}\bar x + {c_2}} \right)^2} = c_1^2\sum\limits_{i = 1}^n {\left( {x_i^2 - 2{x_i}\bar x + {{\bar x}^2}} \right)} + n\left( {c_1^2{{\bar x}^2} + 2{c_1}{c_2}\bar x + c_2^2} \right)\\ = c_1^2 \cdot \sum\limits_{i = 1}^n {x_i^2} - 2c_1^2\bar x\underbrace {\sum\limits_{i = 1}^n {{x_i}} }_{ = n \cdot \bar x} + nc_1^2{{\bar x}^2} + nc_1^2{{\bar x}^2} + 2n{c_1}{c_2}\bar x + nc_2^2\\ = c_1^2\sum\limits_{i = 1}^n {x_i^2} - 2nc_1^2{{\bar x}^2} + 2nc_1^2{{\bar x}^2} + 2n{c_1}{c_2}\bar x + nc_2^2\\ = c_1^2\sum\limits_{i = 1}^n {x_i^2} + 2n{c_1}{c_2}\bar x + nc_2^2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;...\left( 2 \right)\end{array}\) \(\) \(\)

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Clearly, expressions labeled with (1) and (2) are equal, which means that the two initial expressions are equal as well, which proves the given equality.