Question #90052

1 Answer
Dec 10, 2016

Answer:

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

Explanation:

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#