What is the value of $\sum_{k=1}^{60} A_k$ knowing that $A_{n+1}+(-1)^nA_n = 2n-1$ for all $n$?

Question

This sequence meet $A_{n+1}+(-1)^nA_n = 2n-1$ condition. What is the result of the sum from $A_1$ to $A_{60}$?

Answer 1

Notice:

$$A_{n+1}=2n-1-(-1)^nA_n$$ Finding $A_n$ for $n=2,3,4$ in terms of $A_1$ $$A_2=2-1-(-1)A_1\to A_2=A_1+1$$ This implies the common difference $d=1$.

$$A_3=4-1-(1)[A_1+1]=2-A_1$$ $$A_4=6-1-(-1)[2-A_1]=7-A_1$$

This however, implies that $d=5$, a contradiction. So your sequence is not arithmetic.

Answer 2

The condition you have been given can be broken up into an odd condition and an even condition:

$$A_{2n}-A_{2n-1}=2(2n-1)-1$$ and $$A_{2n+1}+A_{2n}=2(2n)-1.$$

Notice that you can break down your sum $$\sum_1^{60} A_n$$ into parts of these forms so long as you $n$ is divisible by four. For example $$\sum_{i-1}^4 = (A_4-A_3)+2(A_3+A_2)-(A_2-A_1)$$ and $$\sum_{i-1}^8 = (A_8-A_7) +2(A_7+A_6) - (A_6-A_5) + (A_4-A_3)+2(A_3+A_2)-(A_2-A_1).$$ Now it's just a matter of using those two formulas above.