It is known that a (n) and B (n) are all arithmetic sequences with tolerance of 1, the first terms are a (1) and B (1), and a (1) + B (1) = 5, a (1), B (1) ∈ positive integers. Let C (n) = a (b (n)), n ∈ positive integers, then the sum of the first ten terms of C (n) is————

It is known that a (n) and B (n) are all arithmetic sequences with tolerance of 1, the first terms are a (1) and B (1), and a (1) + B (1) = 5, a (1), B (1) ∈ positive integers. Let C (n) = a (b (n)), n ∈ positive integers, then the sum of the first ten terms of C (n) is————

Let a (n) = a (1) + n-1. According to the meaning, a (B1) = a (1) + B (1) - 1 = 4
So C (1) = 4. If the tolerance is known to be 1, the sum of the first ten terms of the sequence C (n) = 4 + 5 + 6 + +13=85
Maybe the owner doesn't understand the meaning of the title, so just take B1 in

Who can give me a detailed introduction to the four methods of solving the general term formula of a sequence of numbers? 1. The split term elimination method 2. The undetermined coefficient method 3. The accumulation method 4. It seems to be the cumulative multiplication or something Thank you very much!

One split term elimination method (such as 1 / (1 * 2) + 1 / (2 * 3) +...) +1/n(n+)=1-1/2+1/2-1/3+…… +1 / n-1 / (n + 1) = 1-1 / (n + 1) = n / (n + 1) actually uses the formula: 1 / N (n + 1) = 1 / n-1 / (n + 1), which is the split term