Let f ( x ) be a polynomial of degree 3 such that f(k) = - (2)/(k)...

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$</p><p>$2 = a ( - 2 ) ( - 3 ) ( - 4 ) ( - 5 )$</p><p>$a = \dfrac{1}{60}$</p><p>Put$a = \dfrac{1}{60}$in$( i )$, we get</p><p>$kf ( k ) + 2 = \dfrac{1}{60} ( k - 2 ) ( k - 3 ) ( k - 4 ) ( k - 5 )$</p><p>Now, put$k = 10$</p><p>$10f ( 10 ) + 2 = \dfrac{1}{60} \times 8 \times 7 \times 6 \times 5$</p><p>$10f ( 10 ) = 26$</p><p>So,$52 - 10f ( 10 ) = 26$