Calculate 1**2 + 2**2 + 3**2 + ...

seq -s^2+ 11 |rev| cut -d'+' -f2- | rev | bc
I can't put the last ^2 with seq, so I reverse it to delete the last +N. So for doing sum(N^2) you have to do sum((N+1)^2). Must be a better way.
Sample Output
385

-2
2011-02-10 08:41:14

2 Alternatives + Submit Alt

What Others Think

I like your use of the seq separator, but this time it doesn't quite work. You can replace rev, cut, rev with sed: seq -s^2+ 11 | sed 's/$/^2/' | bc or go back to sed and paste seq 11 | sed 's/$/^2/' | paste -sd+ | bc These also mean that "seq N" gives the sum to N^2.
flatcap · 792 weeks and 4 days ago
With some math python -c 'n=10;print n*(n+1)*(2*n+1)/6'
sarcilav · 792 weeks and 4 days ago
The same idea, but with echo and bc echo 'n=10;n*(n+1)*(2*n+1)/6'|bc
sarcilav · 792 weeks and 4 days ago
Thanks flatcap! I shoud have think o adding a trailing ^2 :S.
rubenmoran · 792 weeks and 4 days ago

What do you think?

Any thoughts on this command? Does it work on your machine? Can you do the same thing with only 14 characters?

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