An interesting one

Find ax5 + by5 where a,b,x,y are real numbers satisfying

ax+by = 3

ax2+by2 = 7

ax3+by3 = 16

ax4+by4 = 42

2 Answers

1
pritishmasti ............... ·

ans=20
using the identity - (x+y)(axn + byn) - xy(axn-1 + byn-1) = axn+1 + yn+1

1
Dipanjan Das ·

ax+by =3
=>ax2+bxy=3x

ax2+by2=7

from these two:

by(x-y)=3x-7......................................................(1)

from the rest of the equations:

by2(x-y)=7x-16....................................................(2)
and ,

by3(x-y)=16x-42...................................................(3)

or, y=7x-163x-7=16x-427x-16.

or, x2+14x-38=0...................................................(4)

Now, we do not need to solve this equation...

Notice that whatever we had done in case if x could also have been done with y
i.e.
we can get equations:

ax(y-x)=3y-7......................................................(1')

ax2(y-x)=7y-16....................................................(2')
and ,

ax3(y-x)=16y-42...................................................(3')

or, x=7y-163y-7=16y-427y-16.

or, y2+14y-38=0...................................................(4')

So, x and y are roots of the equation t2+14t-38=0

x+y=-14
xy =-38

Now utilizing the fact that

(x+y)(ax4+by4)=ax5+by5+xy(ax3+by3)

ax5+by5=38*16 - 14*42= 20

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