Summation of seriesAQA A-Level Further Maths: Revision notes

These results are used for sums that start at $r=1$: $\sum_{r=1}^{n}1=n,\qquad \sum_{r=1}^{n}r=\frac{n(n+1)}{2},\qquad \sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}{6},\qquad \sum_{r=1}^{n}r^3=\frac{n^2(n+1)^2}{4}.$ Note that $\sum r^3=\left(\sum r\right)^2$. Check them with small values: for $n=3$, $\sum r^2=1+4+9=14=\frac{3\times4\times7}{6}$.

To sum a series whose general term is a polynomial in $r$, expand it into powers of $r$, split the sum, take out constant factors and use the standard results. Example: $\sum_{r=1}^{n}r(r+2)=\sum r^2+2\sum r=\frac{n(n+1)(2n+1)}{6}+n(n+1)=\frac{n(n+1)(2n+7)}{6}$. Example: $\sum_{r=1}^{n}r(r^2-3)=\sum r^3-3\sum r=\frac{n^2(n+1)^2}{4}-\frac{3n(n+1)}{2}=\frac14n(n+1)(n-2)(n+3)$. Take out the common factor $n(n+1)$ first, then simplify what is left in the bracket.

A sum that does not start at $r=1$ is a difference of two sums from $r=1$: $\sum_{r=a}^{b}f(r)=\sum_{r=1}^{b}f(r)-\sum_{r=1}^{a-1}f(r).$ Example: $\sum_{r=11}^{20}r(r+2)=S_{20}-S_{10}=3290-495=2795$. The subtracted sum stops one term before the start of the range, at $a-1$.

If the general term can be written as $f(r)-f(r-1)$ (or $f(r+1)-f(r)$), most terms cancel in pairs when the sum is written out. This is the method of differences: $\sum_{r=1}^{n}\left[f(r)-f(r-1)\right]=f(n)-f(0).$ Write out the first few and last few terms to see what survives. In the AS course the difference form is given (or shown by combining fractions or factorising), so no partial fractions are needed. Example: $f(r)=r(r+1)(r+2)$ gives $f(r)-f(r-1)=3r(r+1)$, so $3\sum_{r=1}^{n}r(r+1)=f(n)-f(0)=n(n+1)(n+2)$ and $\sum_{r=1}^{n}r(r+1)=\frac{n(n+1)(n+2)}{3}$.

If the pattern is $\frac1r-\frac1{r+2}$, terms cancel two steps apart, so two terms survive at each end. Example: $\frac1{r(r+2)}=\frac12\left(\frac1r-\frac1{r+2}\right)$, which can be checked by combining the fractions. Then $\sum_{r=1}^{n}\frac1{r(r+2)}=\frac12\left(1+\frac12-\frac1{n+1}-\frac1{n+2}\right)=\frac34-\frac{2n+3}{2(n+1)(n+2)}.$ Remember the factor $\frac12$, and keep the signs of the last two terms.