If 2 tan 2 θ - 5 sec θ = 1 has exactly 7 solutions in the interval 0 , n π 2 , for the least value of n ∈ N then ∑ k = 1 n k 2 k is equal to :

Step-by-step solution

Given, 2 tan 2 θ - 5 secθ - 1 = 0 ⇒ 2 sec 2 θ - 1 - 5 secθ - 1 = 0 ⇒ 2 sec 2 θ - 5 secθ - 3 = 0 ⇒ 2 sec 2 θ - 6 secθ + secθ - 3 = 0 ⇒ 2 secθ sec θ - 3 + 1 secθ - 3 = 0 ⇒ ( 2 secθ + 1 ) ( secθ - 3 ) = 0 ⇒ secθ = - 1 2 , 3 ⇒ cosθ = - 2 , 1 3 We know that, - 1 ≤ cosθ ≤1 ⇒ cosθ = 1 3 For 7 solutions n = 13 So, ∑ k = 1 13 k 2 k = S (say) ⇒ S = 1 2 + 2 2 2 + 3 2 3 + … . + 13 2 13 ⇒ 1 2 S = 1 2 2 + 1 2 3 + … . . + 12 2 13 + 13 2 14 ⇒ S - S 2 = 1 2 + 2 2 2 + 3 2 3 + … . + 13 2 13 - 1 2 2 + 1 2 3 + … . . + 12 2 13 + 13 2 14