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Nonn, easy, nice. (Posted on 2022-11-23) Difficulty: 3 of 5

To prove:

The sum of the first n even squares, less the sum of the first n squares, is equal to the nth square, plus the nth cube, plus the nth triangular number.

The sum of the first n odd squares, less the sum of the first n squares, is equal to the nth cube, less the nth triangular number.

No Solution Yet Submitted by broll    
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Some Thoughts computer exploration | Comment 3 of 5 |
sum1=0; sum2=0;tr=0;
for n=1:15
   sum1=sum1+3*n^2;
   tr=tr+n;
   sum2=n^2+n^3+tr;
   disp([sum1 sum2])
end
disp(' ')
sum1=0; sum2=0;tr=0;
for n=1:15
   sum1=sum1+(2*n-1)^2-n^2;
   tr=tr+n;
   sum2= n^3-tr;
   disp([sum1 sum2])
end

finds equal compares for both parts:

>> nonnEasyNice
     3     3
    15    15
    42    42
    90    90
   165   165
   273   273
   420   420
   612   612
   855   855
  1155  1155        
  1518  1518      
  1950  1950      
  2457  2457      
  3045  3045      
  3720  3720      
 
     0     0
     5     5
    21    21
    54    54
   110   110
   195   195
   315   315
   476   476
   684   684
   945   945
  1265  1265
  1650  1650
  2106  2106
  2639  2639
  3255  3255



They appear to be A059270 and A160378 in the OEIS, which describes each as nonn and easy.


  Posted by Charlie on 2022-11-23 11:20:15
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