If 1 1 0 1 1 2 0 1 1 3 0 1 … . 1 n - 1 0 1 = 1 78 0 1 , then the inverse of 1 n 0 1 is:

Step-by-step solution

Given 1 1 0 1 1 2 0 1 1 3 0 1 . . . . 1 n - 1 0 1 = 1 78 0 1 And, we have 1 1 0 1 1 2 0 1 = 1 3 0 1 = 1 1 + 2 0 1 Also, 1 1 0 1 1 2 0 1 1 3 0 1 = 1 3 0 1 1 3 0 1 = 1 6 0 1 = 1 1 + 2 + 3 0 1 ⇒ 1 1 0 1 1 2 0 1 1 3 0 1 . . . . 1 n - 1 0 1 = 1 1 + 2 + 3 + . . . . + n - 1 0 1 ⇒ 1 1 + 2 + 3 + . . . . + n - 1 0 1 = 1 78 0 1 Using the sum of first n natural numbers i.e. 1 + 2 + 3 + . . . + n = n n + 1 2 , we get n ( n - 1 ) 2 = 78 ⇒ n 2 - n - 156 = 0 ⇒ n - 13 n + 12 = 0 ⇒ n = 13 or n = - 12 (r