Practice section 5 of 17

Evaluate: (i) $1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2+9^2+10^2+11^2+12^2+13^2+14^2+15^2$; (ii) $\sum_{r=1}^{20}(3-2r)$; (iii) $\sum_{r=0}^{15}2^r$.

Find the least positive integer $n$ for which $\sum_{r=1}^{n}1/r>4$.

Add or subtract the missing terms to make each equation valid: (a) $\sum_{r=2}^{n}r^2=\sum_{r=1}^{n}r^2-\square$; (b) $\sum_{r=5}^{10}1/(r+1)=\sum_{r=2}^{10}1/(r+1)-\square$; (c) $\sum_{n=1}^{N}2/n=\sum_{n=3}^{N}2/n+\square$.

Given $\displaystyle\sum_{n=1}^{N}\frac1{n(n+1)(n+2)}=\frac14-\frac1{2(N+1)}+\frac1{2(N+2)}$, find $\displaystyle\sum_{n=3}^{N}\frac1{n(n-1)(n-2)}$ for integer $N\ge3$.

Evaluate (a) $\sum_{r=-2}^{7}5$; (b) $\sum_{r=3}^{10}(-2)$.

In the source story, Gauss rapidly finds $1+2+3+\cdots+98+99+100=5050$. How does pairing terms explain the result? What pattern makes the calculation quick?