ax + by = 3 ax^2 + by^2 = 7 ax^3 + by^3 = 16 ax^4 + by^4 = 42 ax^5 + by^5 = ?


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Question Number 71903 by naka3546 last updated on 21/Oct/19

$a x + b y = 3$ ${a x}^{2} + {b y}^{2} = 7$ ${a x}^{3} + {b y}^{3} = 16$ ${a x}^{4} + {b y}^{4} = 42$ ${a x}^{5} + {b y}^{5} = ?$
	Answered by mind is power last updated on 21/Oct/19
	$(x+y)(ax^k +by^k )=ax^(k+1) +by^(k+1) +xy(ax^(k−1) +by^(k−1) ) ⇒ { (((x+y).3=7+xy(a+b)..1)),((7(x+y)=16+xy(3)..2)) :} 16(x+y)=42+xy.(7)...3 2&3 s=x+y and p=xy⇒ 7s=16+3p 16s=42+7p 16(((16+3p)/7))=42+7p ⇒256+48p=294+49p p=−38=xy s=((16−114)/7)=((−98)/7)=−14=x+y (x+y).(ax^4 +by^4 )=ax^5 +by^5 +xy(ax^3 +by^3 ) ⇒−14(42)=ax^5 +by^5 −38(16) ⇒ax^5 +by^5 =38.16−14.42=20$
	$(x + y) ({ax}^{k} + {by}^{k}) = {ax}^{k + 1} + {by}^{k + 1} + xy ({ax}^{k - 1} + {by}^{k - 1})$ $\Rightarrow {\begin{cases} (x + y) . 3 = 7 + xy (a + b) . . 1 \\ 7 (x + y) = 16 + xy (3) . . 2 \end{cases}$ $16 (x + y) = 42 + xy . (7) . . . 3$ $2 & 3 s = x + y and p = xy \Rightarrow$ $7 s = 16 + 3 p$ $16 s = 42 + 7 p$ $16 (\frac{16 + 3 p}{7}) = 42 + 7 p$ $\Rightarrow 256 + 48 p = 294 + 49 p$ $p = - 38 = xy$ $s = \frac{16 - 114}{7} = \frac{- 98}{7} = - 14 = x + y$ $(x + y) . ({ax}^{4} + {by}^{4}) = {ax}^{5} + {by}^{5} + xy ({ax}^{3} + {by}^{3})$ $\Rightarrow - 14 (42) = {ax}^{5} + {by}^{5} - 38 (16)$ $\Rightarrow {ax}^{5} + {by}^{5} = 38.16 - 14.42 = 20$
	Commented by MJS last updated on 22/Oct/19

	$great!$
	Commented by mind is power last updated on 23/Oct/19

	$thanx sir$
	Answered by MJS last updated on 21/Oct/19

	$solve (1) for y, (2) for b, (3) for a; these are$ $all linear, then (4) for x (quadratic)$ $\Rightarrow$ $x = - 7 \pm \sqrt{87}$ $y = - 7 \mp \sqrt{87}$ $a = \frac{49}{76} \pm \frac{457 \sqrt{87}}{6612}$ $b = \frac{49}{76} \mp \frac{457 \sqrt{87}}{6612}$ ${a x}^{5} + {b y}^{5} = 20$
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