Sigma notation and arithmetic seriesAQA A-Level Maths: Revision notes

The symbol $\sum$ means 'sum of'. In $\sum_{r=1}^{n}u_r$, the limits show that $r$ runs from $1$ to $n$, and each value is substituted into the expression $u_r$. Example: $\sum_{r=1}^{4}(3r+2)=5+8+11+14=38$. Useful rules: $\sum_{r=1}^{n}(u_r+v_r)=\sum u_r+\sum v_r$, $\sum_{r=1}^{n}k=nk$ for a constant $k$, and a sum from $r=m$ to $n$ is $S_n-S_{m-1}$, where $S_n$ is the sum from $r=1$. Example: $\sum_{r=10}^{30}=S_{30}-S_9$, which has $30-10+1=21$ terms.

In an arithmetic sequence each term is the previous term plus a constant common difference $d$. With first term $a$: $u_n=a+(n-1)d.$ For $a=7$, $d=4$: $u_{20}=7+19\times4=83$. The sequence $5,8,11,14,\dots$ has $u_r=3r+2$, so $\sum(3r+2)$ is an arithmetic series. You can find $a$ and $d$ from two terms by simultaneous equations: $u_3=17$ and $u_{10}=45$ give $a+2d=17$ and $a+9d=45$, so $d=4$, $a=9$.

The sum of the first $n$ terms of an arithmetic series is $S_n=\frac n2\left(2a+(n-1)d\right)=\frac n2(a+l),$ where $l$ is the last term. Use the second form when the last term is known. For $a=7$, $d=4$, $n=20$: $S_{20}=10(14+76)=900$. In sigma notation: $\sum_{r=1}^{n}(3r+2)=\frac n2\left(5+3n+2\right)=\frac{n(3n+7)}{2}$.

When the sum is given, form an equation in $n$. For $a=9$, $d=4$ and $S_n=1425$: $\frac n2\left(18+4(n-1)\right)=2n^2+7n=1425\Rightarrow(n-25)(2n+57)=0.$ Reject the negative root, since $n$ is a positive integer: $n=25$. For an inequality such as $S_n>5000$ with $S_n=2n^2+5n$, solve the quadratic equation, then round up to the next whole number: $n=48.77\dots$ gives $n=49$. Check by evaluating $S_{48}=4848$ and $S_{49}=5047$.

Given $S_8=148$ and $S_{12}=342$: $4(2a+7d)=148\Rightarrow2a+7d=37,\qquad6(2a+11d)=342\Rightarrow2a+11d=57.$ Subtract: $4d=20$, so $d=5$ and $a=1$. Then $u_r=5r-4$ and $\sum_{r=1}^{n}(5r-4)=\frac{n(5n-3)}{2}$, so $\sum_{r=11}^{20}(5r-4)=S_{20}-S_{10}=970-235=735$. Show every step: substitute into the formula, simplify, solve.