for all n>=1, prove by induction fibonachi(3n) is even Solution base case n=1 f(3) = f(1) + f(2) = 1 +1 = 2 is even let assume that given is true for n=k thus f(3k) is even now we have to prove that this is true for n = k+1 thus consider f(3(k+1) = f(3k+3) = f(3k+1) + f(3k+2) f(3k+1) + f(3k+1) + f(3k) = f(3k) + 2 * ( f(3k+1) since f(3k) is already even we will write that as 2K thus f(3k+3) = 2K + 2*f(3k+1) = 2( K+f(3k+1)) which is multiple of 2 thus even thus according to mathematical indcution f(3n) is even thus proved..