If Σ k = 1 ∞ 1 k + 2 k + k k + 2 = a + b c , where a , b , c ∈ N and a , b , c ∈ [ 1 , 15 ] , then a + b + c is equal to

Step-by-step solution

Here, general term T k = 1 ( k + 2 ) k + k k + 2 Rationalising the expression, we get, ⇒ T k = k + 2 k - k k + 2 k k + 2 2 - k 2 k + 2 ⇒ T k = ( k + 2 ) k − k k + 2 2 k ( k + 2 ) = 1 2 ( 1 k − 1 k + 2 ) Now, T 1 = 1 2 ( 1 1 − 1 3 ) , T 2 = 1 2 ( 1 2 − 1 4 ) , T 3 = 1 2 1 3 - 1 5 and so on T k = 1 2 ( 1 k − 1 k + 2 ) Adding all the terms, we get, T 1 + T 2 + . .. T k = 1 2 ( 1 + 1 2 − 1 k + 1 − 1 k + 2 ) Now, if k → ∞ ∑ k = 1 ∞ T k = 1 2 ( 1 + 1 2 − 0 − 0 ) = 1 + 2 2 2 1 + 2 8 = 2 + 4 16 As, c ∈ [ 1 , 15 ] ⇒ c = 8 ,