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

mathcounterexamples.net
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Shubham Kumar
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  • This has nothing to do with arithmetic progressions. – Thomas Andrews Jun 09 '18 at 15:30
  • @ThomasAndrews $1+2+...+n$, $2+4+6+...+2n$ are arithmetic progressions, and they are involved in the problem. "This has nothing to do with arithmetic progressions" is false. Maybe you want to say "this has more to do with recurrence equations", or "with difference equations", ... –  Jun 09 '18 at 15:33

2 Answers2

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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.

user1390
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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.

Rhys Hughes
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