Question #90052

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Answer 1 · 2016-12-10T14:35:56

$x^n+1/a+y^n+1/b=2((ab)/(a+b))^n+1/a+1/b$

Assuming that the correct formulation is

If $x/a+y/b=1 and x^2/a+y^2/b=(ab)/(a+b)$

then $x^n+1/a+y^n+1/b=$

From

${(x/a+y/b=1),(x^2/a+y^2/b=(ab)/(a+b)):}$

solving for $x,y$ we obtain

$x=y=(ab)/(a+b)$

To obtain this result we solve for $X=x/a,Y=y/b$

${(X+Y=1),(aX^2+bY^2=(ab)/(a+b)):}$

$aX^2+b(1-X)^2=(ab)/(a+b)$ and solving for $X$

$(a+b)X^2-2bX+b=(ab)/(a+b)$ or

$X^2-(2b)/(a+b)X+(b/(a+b))^2$ or

$(X-b/(a+b))^2=0$

then $X=b/(a+b)$ and analogously $Y=a/(a+b)$

so $x=y=(ab)/(a+b)$

Finally we have

$x^n+1/a+y^n+1/b=2((ab)/(a+b))^n+1/a+1/b$