An interesting problem

ax+by=3 ; ax^2+by^2=7 ; ax^3+by^3=16 ; ax^4+bx^4=42 ; Then find ax^5+bx^5 given a,b,x,y are real nos.

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Solution : Note that , for any $n >0 , (ax^n+by^n)(x+y)-xy(ax^{n-1}+by^{n-1}) = ax^{n+1}+by^{n+1}$

So now For $n=2$ we get $7(x+y)-3xy=16$ and for $n=3$ we get $16(x+y)-7xy=42$ .
Solving gives , $x+y=-14 , xy=-38$

now again using the identity ,

$ax^5+by^5 = (42)(-14)-(16)(-38) = \boxed{20}$

Shivang Jindal - 8 years, 4 months ago

Is it $ax^4 + by^4 = 42, ax^5 + by^5$ ? I think you typed wrongly.

Zi Song Yeoh - 8 years, 4 months ago

If it is, then the answer should be $20$ .

Zi Song Yeoh - 8 years, 4 months ago

a/(ax - 1) + b/(bx - 1) = a + b . Find x .

Himanshu Singh - 6 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$