Chapter 11.4, Problem 34WE

To determine

To calculate:

Sigma notation of the following:

  $\log \left(1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\right)$

Answer to Problem 34WE

The sigma notation for $\log \left(1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\right)$ is ${\displaystyle \sum _{n=1}^6\log n}$ .

Explanation of Solution

Given information:

  $\log \left(1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\right)$

Calculation:

Need to solve the given equation first:

  $\begin{array}{l}\log \left(1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\right)=\log 1+\log 2+\log 3+\log 4+\log 5+\log 6\\ \left[\because \log \left(xy\right)=\log x+\log y\right]\end{array}$

From above equation:

The first term is $\log 1$ and the last term is $\log 6$ . Each term is logarithm of power of index $n$ .

So, it becomes: $\log n$

The value of the upper limit is calculated by using the last term as:

  $\begin{array}{l}\log n=\log 6\\ n=6\end{array}$

The value of the lower limit is calculated by using the last term as:

  $\begin{array}{l}\log n=\log 1\\ n=1\end{array}$

So, the lower and upper limit is $1$ and $6$ . Now, the sigma notation for the given series is expressed as:

  $\begin{array}{l}\log 1+\log 2+\log 3+\log 4+\log 5+\log 6={\displaystyle \sum _{n=1}^6\log n}\\ \log \left(1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\right)={\displaystyle \sum _{n=1}^6\log n}\end{array}$