PE[28]: sum of the numbers on the diagonals in a 1001 by 1001 spiral

FreddyMusic

It seems your guys are hungry ?

( Wolves, Warriors ) ?

jimogsh

观察发现，每一圈上的数字的平均值都是该圈上位于1正左边的数字，1，6，19，40……对这串数字进行观察发现，其逐项之差成等差数列，公差为8。

1. In[15]:= RSolve[{a[n + 1] == 2 a[n] - a[n - 1] + 8, a[1] == 1,
   
2.   a[2] == 6}, a[n], n]
   
3. 
   
4. Out[15]= {{a[n] -> 4 - 7 n + 4 n^2}}
   
5. 
   
6. In[19]:= a = 4 #^2 - 7 # + 4 &; Total[4 Table[a[n], {n, 2, 501}]] + 1
   
7. 
   
8. Out[19]= 669171001

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FreddyMusic

Alternative shorter form

1. 
   
2. Total[4 Table[FindSequenceFunction[{1, 6, 19, 40}, n], {n, 1, 501}]]
   

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jimogsh

Alternative shorter form
Total[4 Table[FindSequenceFunction[{1, 6, 19, 40}, n], {n, 1, 501}]]

FreddyMusic 发表于 2009-11-21 21:44

这个把中间的1算了4次，应该再减3，就是答案了。