Given:

**
ax+by=7 **

ax^2+by^2=49

ax^3+by^3=133

ax^4+by^4=406

**
** Evaluate:

** **

** P=2014*(x+y-xy)+100(a+b). **

**
Source: AMO Nepal 2018**

Lets call the equations E(n) for ax^n + by^n = ##.

Then three consecutive equations can be related with E(n+1) = E(n)*(x+y) - E(n-1)*(xy)

So lets plug in the right-hand side values for E() and create a couple of equations (n=2 and n=3 cases):

133 = 49*(x+y) - 7*(xy)

406 = 133*(x+y) - 49*(xy)

Treat this as a system in x+y and xy, and solve to get x+y=5/2 and xy=-3/2

Now a+b is just E(0), so construct one more equation with E(), using n=1:

49 = 7*(x+y) - (a+b)*(xy)

Simply substitute known values for x+y and xy and the simplify to get a+b=21.

Finally P = 2014*(x+y-xy)+100(a+b) = 2014*(5/2+3/2)+100*21 = **10156**.