Create n x 2n "mirror" matrix of this type:
For n = 2
m = [ 1 2 2 1
1 2 2 1 ]
For n = 3
m = [ 1 2 3 3 2 1
1 2 3 3 2 1
1 2 3 3 2 1 ]
This problem was fun. I was able to find a better way to solve beat my own solution. Not the best way though. But lot better.
Sum of first n terms of a harmonic progression
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Return 'on' or 'off'
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Sum of series VII
Sum of series I
Sum of series VI
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