Chapter 11.5, Problem 23WE

To determine

To calculate:

The sum value for the following condition:

Positive three digit odd integers

Answer to Problem 23WE

The sum value of series of positive three digit odd numbers is $247500$ .

Explanation of Solution

Given information:

  $t=4hrs.$

Calculation:

The positive three digit integers start from $100$ and ends on $999$ . But need to define the series of odd numbers:

The first three digit odd integer is $101$ and last the last three odd integer is $999$ between $100$ and $999$ .

So, the series is defined as: $101,103,105,\mathrm{...},999$

Consider,

  $\begin{array}{l}{a}_1=101\\ a_2=103\\ a_3=105\end{array}$

Now, calculate the difference in terms:

  $\begin{array}{l}d=a_2-a_1=103-101=2\\ d=a_3-a_2=105-103=2\end{array}$

Here, each value exceeds by two, so the formula becomes:

  $\begin{array}{l}{a}_n=a_1+\left(n-1\right)d\\ a_n=101+\left(n-1\right)\left(2\right)\\ a_n=101+2n-2\\ a_n=99+2n\end{array}$

Now, calculate the lower and upper limit of the summation notation:

Lower limit:

  $\begin{array}{l}{a}_1=a_1+\left(n-1\right)d\\ 101=101+\left(n-1\right)\left(2\right)\\ n-1=0\\ n=1\end{array}$

And upper limit is:

  $\begin{array}{l}{a}_n=a_1+\left(n-1\right)d\\ 999=101+\left(n-1\right)\left(2\right)\\ 2n-2=898\\ 2n=900\\ n=450\end{array}$

Hence, the summation notation of the series of three digit odd numbers is ${\displaystyle \sum _{n=1}^{450}99+2n}$

Now, by using the summation calculator the sum value of series of positive three digit odd numbers is ${\displaystyle \sum _{n=1}^{450}99+2n}=247500$ .