Activity 52.
For each sum written in sigma notation, write the sum long-hand and evaluate the sum to find its value. For each sum written in expanded form, write the sum in sigma notation.
(a)
\(\displaystyle \sum_{k=1}^{5} (k^2 + 2)\)
Hint.
Observe that when \(k = 1\text{,}\) \(k^2 + 2 = 1^2 + 2 = 3\text{.}\) This is the first term in the sum.
Answer.
\(65 \)
Solution.
\begin{align*}
\sum_{k=1}^{5} (k^2 + 2) =\mathstrut \amp (1^2 + 2) + (2^2 + 2) + (3^2 + 2) + (4^2 + 2) + (5^2 + 2)\\
=\mathstrut \amp 3 + 6 + 11 + 18 + 27\\
=\mathstrut \amp 65
\end{align*}
(b)
\(\displaystyle \sum_{i=3}^{6} (2i-1)\)
Hint.
Note that this sum starts at \(i = 3\text{.}\)
Answer.
\(32 \)
Solution.
\begin{align*}
\sum_{i=3}^{6} (2i-1) =\mathstrut \amp (2 \cdot 3 - 1) + (2 \cdot 4- 1) + (2 \cdot 5 - 1) + (2 \cdot 6 - 1)\\
=\mathstrut \amp 5 + 7 + 9 + 11\\
=\mathstrut \amp 32
\end{align*}
(c)
\(3 + 7 + 11 + 15 + \cdots + 27\)
Hint.
Since the terms in the sum increase by 4, try a function \(f(k)\) that somehow involves \(4k\text{.}\)
Answer.
\begin{equation*}
3 + 7 + 11 + 15 + \cdots + 27 = \sum_{k=1}^{7} 4k-1\text{.}
\end{equation*}
Solution.
Observe that each term in the sum
\begin{equation*}
3 + 7 + 11 + 15 + \cdots + 27
\end{equation*}
differs from the previous term by 4. If we view \(4\) as \(4 = 4 \cdot 1 - 1\) and \(7\) as \(7 = 4 \cdot 2 - 1\text{,}\) we see that the pattern may be represented through the function \(f(k) = 4k-1\text{,}\) so that
\begin{equation*}
3 + 7 + 11 + 15 + \cdots + 27 = \sum_{k=1}^{7} 4k-1\text{.}
\end{equation*}
We note that \(k=7\) is the end value of the index since \(4 \cdot 7 = 28\text{.}\)
(d)
\(4 + 8 + 16 + 32 + \cdots + 256\)
Hint.
What pattern do you observe in the terms of the sum?
Answer.
\begin{equation*}
4 + 8 + 16 + 32 + \cdots + 256 = \sum_{i=2}^{8} 2^i\text{.}
\end{equation*}
Solution.
The sum \(4 + 8 + 16 + 32 + \cdots + 256\) is a sum of powers of \(2\text{,}\) which we can express in sigma notation as
\begin{equation*}
4 + 8 + 16 + 32 + \cdots + 256 = \sum_{i=2}^{8} 2^i\text{.}
\end{equation*}
(e)
\(\displaystyle \sum_{i=1}^{6} \frac{1}{2^i}\)
Hint.
Write every term in the sum as a fraction with denominator \(2^6 = 64\text{.}\)
Answer.
\begin{equation*}
\sum_{i=1}^{6} \frac{1}{2^i} = \frac{63}{64}\text{.}
\end{equation*}
Solution.
\begin{align*}
\sum_{i=1}^{6} \frac{1}{2^i} =\mathstrut \amp \frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{2^6}\\
=\mathstrut \amp \frac{32}{64} + \frac{16}{64} + \frac{8}{64} + \frac{4}{64} + \frac{2}{64} + \frac{1}{64}\\
=\mathstrut \amp \frac{63}{64}\text{.}
\end{align*}
