Guess a particular solution I(n) which is in the form of an integral with lower and upper limits 0 and 1 respectively, to the difference equation,
I(n+1) + I(n) = 1/(3n+1) for all n>=0
Hence, calculate the infinite series:
∑[k=0 to infinity] (-1)^k /(3k+1)
[Hint: Integrate[from 0 to 1] x^(3n) dx = 1/(3n+1)]
Integration to series
2010-07-20 7:11 am
回答 (1)
2010-07-21 3:50 am
✔ 最佳答案
Please see the following:圖片參考：http://img821.imageshack.us/img821/4031/36733347.png
2010-07-20 19:50:50 補充：
http://img821.imageshack.us/img821/4031/36733347.png
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