If b is very small as compared to the value of a , so that the cube and other higher powers of b a can be neglected in the identity 1 a - b + 1 a - 2 b + 1 a - 3 b + … . + 1 a - n b = α n + β n 2 + γ n 3 then the value of γ is :

Step-by-step solution

Given, 1 a - b + 1 a - 2 b + 1 a - 3 b + … . + 1 a - n b = a - b - 1 + a - 2 b - 1 + … . + a - n b - 1 = 1 a ∑ r = 1 n 1 - r b a - 1 = 1 a ∑ r = 1 n 1 + r b a + r 2 b 2 a 2 + terms   to   be   neglected = 1 a n + n ( n + 1 ) 2 · b a + n ( n + 1 ) ( 2 n + 1 ) 6 · b 2 a 2 = 1 a n + n 2 + n 2 · b a + 2 n 3 + 3 n 2 + n 6 · b 2 a 2 So γ = b 2 3 a 3