The usual Fibonacci numbers are given by the recurrence: Solution a) put n=1 we get F1=F3-1 -----------1 we know F3=1+0+1=2 substitute back F3 in equation 1 we get F1=1 which is true now put n=i we get F1+F2+F3+...Fi=F(i+2)-1 now the values of F1 F2 F3 ARE DERIVED FROM DEFINITION OF FIBONACCI F1=1 F2=1 F3=2 F4=3 F5=5 SO WE GET 1+1+2+3+4+.....=F(i+2)-1 we get 1+ i(i+1)/2=F(i+1)+F(i) 1+ i(i+1)/2=2F(i)+F(i-1) but 2F(i)+F(i-1)=1+ i(i+1)/2 therefore lhs=rhs b) put n=1 first we get lhs =rhs now put n=i F2i=F(i+1)F(i-1)+ -1^i+1 substitute the values by definition and take the sum of squares of first n natural numbers by definition of the lhs part and the fibonacci definition on rhs both are equal hope it helps.