[tex]\\\sum_{n=1}^{n=4}(\frac12n)^5=(\frac12)^5+1+(\frac32)^5+2^5=\frac{1}{32}+1+\frac{243}{32}+32= \\33+\frac{244}{32}=33+\frac{61}{8}=33+7\frac58=40\frac58[/tex]
