AI HL · 12.1
0 of 120 skills done
Series & Sigma Notation
12.1.4
Template key, then Σ
Evaluate \(\displaystyle\sum_{r=1}^{5} (3r + 1)\).
Have a go first. Then open the working one line at a time.
\(4 + 7 + 10 + 13 + 16\)
Put \(r = 1, 2, 3, 4, 5\) into \(3r + 1\).
\(50\)
Five terms. It is an arithmetic series with \(d = 3\).
Evaluate \(\displaystyle\sum_{r=1}^{15} 4\).
Have a go first. Then open the working one line at a time.
\(r = 1,\ 2,\ \ldots,\ 15\)
15 values of \(r\), so 15 terms.
\(4 + 4 + \ldots + 4\)
The expression does not depend on \(r\), so each term is 4.
\(15 \times 4 = 60\)
An arithmetic series with \(d = 0\) gives the same answer.
Evaluate \(\displaystyle\sum_{r=4}^{12} (2r + 5)\).
Have a go first. Then open the working one line at a time.
\(r = 4\) gives 13. \(r = 12\) gives 29.
The terms go up by 2: an arithmetic series.
\(12 - 4 + 1 = 9\)
Count both ends.
\(S_9 = \frac{9}{2}(13 + 29) = 189\)
\(S_n = \frac{n}{2}(u_1 + u_n)\).
Evaluate \(\displaystyle\sum_{r=1}^{6} 3 \times 2^r\).
Have a go first. Then open the working one line at a time.
\(6 + 12 + 24 + \ldots\)
Each term is double the one before: geometric with \(u_1 = 6\), \(r = 2\), \(n = 6\).
\(S_6 = \frac{6(2^6 - 1)}{2 - 1} = 378\)
\(S_n = \frac{u_1(r^n - 1)}{r - 1}\).
Find the smallest value of \(n\) for which \(\displaystyle\sum_{r=1}^{n} 2 \times 3^{r-1} > 1000\).
Have a go first. Then open the working one line at a time.
\(2 + 6 + 18 + \ldots\)
Geometric with \(u_1 = 2\) and \(r = 3\).
\(S_n = \frac{2(3^n - 1)}{3 - 1} = 3^n - 1\)
Simplify before you compare.
\(S_6 = 728\), \(S_7 = 2186\)
Use a table on the calculator. 7 terms is the first total above 1000.
Write \(7 + 11 + 15 + \ldots + 47\) in sigma notation, and find its sum.
Have a go first. Then open the working one line at a time.
\(u_r = 7 + (r-1)(4) = 4r + 3\)
Arithmetic with \(u_1 = 7\), \(d = 4\).
\(4r + 3 = 47\) gives \(r = 11\)
So there are 11 terms.
\(\displaystyle\sum_{r=1}^{11} (4r + 3)\)
Start at \(r = 1\), stop at \(r = 11\).
\(S_{11} = \frac{11}{2}(7 + 47) = 297\)
First term 7, last term 47, 11 terms.