Summary
n∑i=1c=nc∑i=1nc=nc
n∑i=1cxi=cn∑i=1xi∑i=1ncxi=c∑i=1nxi
n∑i=1(xi+yi)=n∑i=1xi+n∑i=1yi∑i=1n(xi+yi)=∑i=1nxi+∑i=1nyi
n∑i=1xi=m∑i=1xi+n∑i=m+1xi∑i=1nxi=∑i=1mxi+∑i=m+1nxi