If a = 1 + 2 + 4 + ⋯ ⋯ up to n terms, b = 1 + 3 + 9 + ⋯ ⋯ up to n terms and c = 1 + 5 + 25 + … … up to n terms, then Δ = a 2 b 4 c 2 2 2 2 n 3 n 5 n =

Step-by-step solution

a = 1 + 2 + 4 + 8 + . . . +   upto n terms a = 2 n - 1 b = 1 + 3 + 9 + . . . + upto n terms ⇒ b = 3 n - 1 2 ⇒ 2 b = 3 n - 1 c = 1 + 5 + 25 + . . .   t upto n terms ⇒ c = 5 n - 1 4 4 c = 5 n - 1 ∆ = 2 n - 1 3 n - 1 5 n - 1 2 2 2 2 n 3 n 5 n ∆ = 2 n 3 n 5 n 2 2 2 2 n 3 n 5 n - 2 1 1 1 1 1 1 2 n 3 n 5 n by splitting the determinant into two along the first row. Also, we know that the value of determinant is 0 if there are two identical rows or columns. ∆ = 0