Lets start with a simple identity to find the value of ab:
((a+b)^2 - (a^2+b^2))/2 = ab
Substituting gives us (1^2 - 2)/2 = -1/2 = ab.
Now a second identity is useful
a^n+b^n = [a^(n-1)+b^(n-1)]*(a+b) - [a^(n-2)+b^(n-2)]*(ab)
Now substituting yields a recursive identity for computing the sum of powers of a and b:
a^n+b^n = [a^(n-1)+b^(n-1)] + [a^(n-2)+b^(n-2)]/2
Then starting with initial sums a+b=1 and a^2+b^2=2 and stepping through the recursion yields the sequence
1, 2, 5/2, 7/2, 19/4, 13/2, 71/8, 97/8, 265/16, 181/8, 989/32, ....
The answer is a^11+b^11 = 989/32 = 30 29/32.
Solution
