If x + 1 x = 1 , λ = x 4 0 0 0 + 1 x 4 0 0 0 and μ be the digit at the unit place of the number 2 2 n + 1 , where n ∈ N and n > 1 , then the value of λ + μ  is equal to

Step-by-step solution

∵     x + 1 x = 1 ⇒ x 2 - x + 1 = 0 ⇒ x + ω x + ω 2 = 0 ∴ x = - ω , ⁡ - ω 2 , where ω and ω 2 are complex cube roots of unity if x = - ω , then λ = - ω 4 0 0 0 + 1 - ω 4 0 0 0 = ω 4 0 0 0 + 1 ω 4 0 0 0 ⇒     λ = ω + 1 ω = ω 2 + 1 ω = - ω ω = - 1 ∵     ω 3 = 1 and   1 + ω + 1 ω 2 = 0 and also for x = - ω 2 ,  ⁡ &#955