Let S = _ k=1 ⁵ ^2 ( k 11 ) and P = _ k=1 ⁵ ( k 11 ) . The value of the ratio S P is equal to:

Step-by-step solution

Given, S = _ k=1 ⁵ ^2 ( k 11 ) Using the identity ^2 = 1 - 2 2 , we get: S = 1 2 _ k=1 ⁵ ( 1 - 2k 11 ) = 5 2 - 1 2 _ k=1 ⁵ 2k 11 Let C = 2 11 + 4 11 + 6 11 + 8 11 + 10 11 Multiplying and dividing by 2 11 : C = 2 11 2 11 + + 2 11 10 11 2 11 Using 2 A B = (A+B) - (B-A) : C = ( 3 11 - 11 ) + ( 5 11 - 3 11 ) + + ( 11 11 - 9 11 ) 2 11 This is a telescoping sum that simplifies to: C = - 11 2 11 = - 1 2 Thus, S = 5 2 - 1 2 ( - 1 2 ) = 11 4 Now for the product P = 11 2 11 3 11 4 11 5 11 We know 3 11 = - ( - 3 11 ) = - 8 11