Porisma

Worked solution

Express and evaluate an arithmetic series in two ways

Express 1+4+7+10+⋯+611+4+7+10+\cdots+61 in sigma notation and find its sum.

CJC · Sequences and Series · Example 22

Answer

∑r=121(3r−2)=651\sum_{r=1}^{21}(3r-2)=651.

Compare your reasoning

Find the number of terms

un=1+(n−1)3=61⟹3n−2=61⟹n=21u_n=1+(n-1)3=61\quad\Longrightarrow\quad3n-2=61\quad\Longrightarrow\quad n=21

The source identifies a=1,d=3 and uses the last-term equation.

Method 1: use the AP sum formula

S21=212[2(1)+(21−1)3]=651\begin{aligned}&S_{21}\\ & =\frac{21}{2}[2(1)+(21-1)3]\\ & =651\end{aligned}

The source callout permits AP concepts when a question does not explicitly require sigma rules as the solving method. This question still asks for a sigma representation.

Method 2: apply the sigma rules

1+4+7+10+⋯+61=∑r=121(3r−2)=3∑r=121r−∑r=1212=3[212(1+21)]−2(21)=651.\begin{aligned}1+4+7+10+\cdots+61&=\sum_{r=1}^{21}(3r-2)\\&=3\sum_{r=1}^{21}r-\sum_{r=1}^{21}2\\&=3\left[\frac{21}{2}(1+21)\right]-2(21)\\&=651.\end{aligned}

The source points to the constant-factor and sum/difference rules. Keep the complete natural-number and constant sums.

How did this question go?