Q. 43 - Calculus

Question

Write each expression in Exercises 41–43 in one sigma notation (with some extra terms added to or subtracted from the sum, as necessary). 

$3\sum _{k=2}^{25}{k}^2+2\underset{k=2}{\overset{24}{\sum k}}-\sum _{k=0}^{25}1$

Step-by-Step Solution

Answer

The expression in one sigma notation is: $1872+ \sum _{k=2}^{24}\left(3k^2+2k-1\right)$

Step 1. Given information

The expression is:

$\underset{k=2}{\overset{25}{3\sum }}{k}^2+2\underset{k=2}{\overset{24}{\sum k}}-\sum _{k=0}^{25}1$

Step 2. Write expression into one sigma notation:

$$\begin{gathered} \underset{k=2}{\overset{25}{3\sum }}{k}^2+2\underset{k=2}{\overset{24}{\sum k}}-\sum _{k=0}^{25}1=\left(\sum _{k=2}^{24}3k^2+3{\left(25\right)}^2\right)+2\underset{k=2}{\overset{24}{\sum k}}-\left(1+1+\sum _{k=2}^{24}1+1\right) \\ =\left(\sum _{k=2}^{24}3k^2+1875\right)+\underset{k=2}{\overset{24}{\sum 2k}}-\left(3+\sum _{k=2}^{24}1\right) \\ =1875-3+ \sum _{k=2}^{24}\left(3k^2+2k-1\right) \\ =1872+ \sum _{k=2}^{24}\left(3k^2+2k-1\right) \end{gathered}$$