f(x)+f(x-1)=x^2.If f(55) = 89, what is the value of f(89)? (It is not 55.)
f(56) +f(55) = 56^2
f(56) + 89 = 56^2 = 3136
f(56) = 3047
f(89) +f(88) = 89^2 equation (1)
f(88) +f(87) = 88^2 equation (2)
…………………….. …………
…………………….. ………….
f(56) +f(55) =56^2 …………..
by substracting all the equations from equation (1)
we get:
f(89) -f(55) = 89^2 -88^2 – 87^2 -……..-56^2
f(89) -f(55) = 89^2 -{ 88^2 +87^2 +…..+56^2}
we use the series formula
let S1 =1^2 +2^2 +3^2 +….+ 55^2 = 55(56)(111)/6 =56980
and
let S2 =1^2 +2^2 +3^2 +…..+88^2 = 88(89)(177)/6 =231044
S2 -S1 = 88^2 +87^2 +…+56^2 = 231044 -56980 =177064
hence ,
f(89) -f(55) = 89^2 -177064 = – 169143
f(89) = -168143 +f(55) = -168143 +89 =-169054
(check for errors)
f(x+1) + f(x) = (x+1)²
f(x) + f(x-1) = x²
Subtracting,
f(x+1) – f(x-1) = 2x + 1
So I just considered positive integers for x and treated the cases x odd and x even separately. The result, for x a positive integer, is
f(x) = x(x+1)/2 + 1451 for x even
f(x) = x(x+1)/2 – 1451 for x odd
or more compactly,
f(x) = x(x+1)/2 + 1451*(-1)^x
Then f(89) = 2554
A generalization for x not necessarily an integer (which does satisfy the recursion relation) would be
f(x) = x(x+1)/2 + 1451*cos(? x)
There’s probably an elegant way to solve this…
f(x) + f(x-1) = x^2, i.e. f(x) = x^2 – f(x-1).
Using this to compute successive values I get f(89) = 2554. This is actually the fastest way of doing it if you have a spreadsheet handy. 
Otherwise, note that you get
f(89) = 89^2 – f(88)
= 89^2 – (88^2 – f(87))
= 89^2 – 88^2 + 87^2 – f(86)
= … = 89^2 – 88^2 + 87^2 – 86^2 + … + 57^2 – 56^2 + f(55).
So we have to evaluate 89^2 – 88^2 + 87^2 – 86^2 + … + 57^2 – 56^2
= (2(88) + 1) + (2(86) + 1) + … + (2(56) + 1)
= 2(88 + 86 + … + 56) + 17 (there are 17 terms)
= 4(44 + 43 + … + 28) + 17
If you remember 1 + 2 + … + n = n(n+1)/2 we get
= 4(44(45)/2 – 27(28)/2) + 17
= 4(990 – 378) + 17
= 2465
so f(89) = 2465 + f(55) = 2465 + 89 = 2554.