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.
Solution