For a ∈ R (the set of all real numbers), a ≠ - 1 , lim n → ∞ 1 a + 2 a + … + n a n + 1 a - 1 na + 1 + na + 2 + … + na + n = 1 6 0 Then a =

Step-by-step solution

This numerical can be solved by using the concept of Definite integral as limit of sum. Lim n → ∞ ∑ r a r = 1 n n + 1 a - 1 ∑ r = 1 n ( n a + r ) Divide Numerator and Denominator by n a and rearranging the terms Lim n → ∞ ∑ r = 1 n r n a 1 + 1 n a - 1 ∑ r = 1 n a + r n Now Multiply Numerator and denominator by 1 n Lim n → ∞ ∑ r = 1 n r n a × 1 n 1 + 1 n a - 1 ∑ r = 1 n a + r n × 1 n Now Replace 1 n by d x     and r n by x an