- Thread starter
- #1
- Mar 10, 2012
- 834
The following is identically 0 which can be readily checked by a simple hand calculation.
$\binom{n+1}{k}2^{-n-1} - \binom{n}{k}2^{-n} + \binom{n}{k}*2^{-n-1} - \binom{n}{k-1}2^{-n-1}$
If you enter this in SAGE or Mathematica, using the appropriate script, and use full_simplify() and FullSimplify[] respectively, you will find that both of these softwares show you the result 0.
Can somebody tell me how does a computer handle such expressions symbolically?
$\binom{n+1}{k}2^{-n-1} - \binom{n}{k}2^{-n} + \binom{n}{k}*2^{-n-1} - \binom{n}{k-1}2^{-n-1}$
If you enter this in SAGE or Mathematica, using the appropriate script, and use full_simplify() and FullSimplify[] respectively, you will find that both of these softwares show you the result 0.
Can somebody tell me how does a computer handle such expressions symbolically?