Given that: kf(k)+2=0 for k=2,3,4,5 which means (k−2),(k−3),(k−4),(k−5) are the factors of this expression.
Let kf(k)+2=a(k−2)(k−3)(k−4)(k−5) ....(i)
Putk=0
2=a(−2)(−3)(−4)(−5)
a=601
Put a=601 in (i), we get
kf(k)+2=601(k−2)(k−3)(k−4)(k−5)
Now, put k=10
10f(10)+2=601×8×7×6×5
10f(10)=26
So, 52−10f(10)=26