∑k=1nk=n(n+1)2\displaystyle\sum_{k=1}^{n} k=\frac{n(n+1)}{2}k=1∑nk=2n(n+1)
∑k=1nk2=n(n+1)(2n+1)6\displaystyle\sum_{k=1}^{n} k^2=\frac{n(n+1)(2n+1)}{6}k=1∑nk2=6n(n+1)(2n+1)
∑k=1nk3={n(n+1)2}2\displaystyle\sum_{k=1}^{n} k^3=\Big\{\frac{n(n+1)}{2}\Big\}^2k=1∑nk3={2n(n+1)}2
∑k=1n(ak±bk)=∑k=1nak±∑k=1nbk\displaystyle\sum_{k=1}^{n} (a_k \pm b_k)=\sum_{k=1}^{n} a_k \pm \sum_{k=1}^{n} b_kk=1∑n(ak±bk)=k=1∑nak±k=1∑nbk
∑k=1ncak=c∑k=1nak\displaystyle\sum_{k=1}^{n} ca_k = c\sum_{k=1}^{n} a_kk=1∑ncak=ck=1∑nak
∑k=1nc=cn\displaystyle\sum_{k=1}^{n} c = cnk=1∑nc=cn
