Find the value of
(a) $\displaystyle\sum_{r=1}^{40}(3r-2)$ [2]
(b) $\displaystyle\sum_{r=5}^{12}2^{r}$ [3]
- (a) Recognises an arithmetic series with $a=1,\ d=3$ (or $l=118$) and $n=40$ and uses a correct sum formula, e.g. $\tfrac{40}{2}(1+118)$M1
- $2380$A1
- (b) Identifies a geometric series with first term $2^{5}=32$ and $r=2$ (or writes $\sum_{1}^{12}-\sum_{1}^{4}$)M1
- Uses $S_n=\dfrac{a(r^{n}-1)}{r-1}$ with $n=8$ (or $2^{13}-2-(2^{5}-2)$)M1
- $8160$A1
- Answer: (a) $2380$ (b) $8160$
- Examiner tip: Count the terms from $r=5$ to $r=12$ carefully: there are 8, not 7.