{ ((ax+by=7)),((ax^2 +by^2 =49)),((ax^3 +by^3 =133)),((ax^4 +by^4 =406)) :} ⇒2014x+2014y−100a−100b−2014xy=?


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Question Number 157664 by cortano last updated on 26/Oct/21
${ ((ax+by=7)),((ax^2 +by^2 =49)),((ax^3 +by^3 =133)),((ax^4 +by^4 =406)) :} ⇒2014x+2014y−100a−100b−2014xy=?$
${\begin{cases} a x + b y = 7 \\ {a x}^{2} + {b y}^{2} = 49 \\ {a x}^{3} + {b y}^{3} = 133 \\ {a x}^{4} + {b y}^{4} = 406 \end{cases}$ $\Rightarrow 2014 x + 2014 y - 100 a - 100 b - 2014 x y = ?$
Commented by mr W last updated on 26/Oct/21

$x + y = \frac{7 \times 406 - 49 \times 133}{7 \times 133 - 49^{2}} = \frac{5}{2}$ $x y = \frac{49 \times 406 - 133^{2}}{7 \times 133 - 49^{2}} = - \frac{3}{2}$ $a + b = \frac{7 \times \frac{5}{2} - 49}{- \frac{3}{2}} = 21$ $2014 x + 2014 y - 100 a - 100 b - 2014 x y$ $= 2014 (x + y - x y) - 100 (a + b)$ $= 2014 (\frac{5}{2} + \frac{3}{2}) - 100 \times 21$ $= 2014 \times 4 - 100 \times 21$ $= 5956$
Commented by mr W last updated on 26/Oct/21

$f o r d e t a i l s s e e Q 87533$
Commented by cortano last updated on 26/Oct/21

$a m a z i n g$
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