Question

9Evaluate Σ(3 +4).A=1Ο Α. 247Ο Β. 321Ο C. 859Ο D. 891

Answer 1

891 (option D)

Explanation:$\sum ^9_{i=1}(3i^2\text{ + 4)}$

To evaluate the above, we will find the values of each termat i = 1 - 9.

The we will add the values together

$\begin{gathered} \text{when i = 1} \\ 3i^2+4=3(1)^2\text{ + 4} \\ =\text{ 7} \\ \\ \text{when i = 2} \\ 3i^2+4=3(2)^2\text{ + 4} \\ =\text{ 16} \\ \\ \text{when i = 3} \\ 3i^2+4=3(3)^2\text{ + 4} \\ =\text{ 31} \end{gathered}$$\begin{gathered} \text{when i = 4} \\ 3i^2+4=3(4)^2\text{ + 4} \\ =\text{ }52 \\ \\ \text{when i =5} \\ 3i^2+4=3(5)^2\text{ + 4} \\ =\text{ }79 \\ \\ \text{when i = 6} \\ 3i^2+4=3(6)^2\text{ + 4} \\ =\text{ 112} \end{gathered}$$\begin{gathered} \text{when i = 7} \\ 3i^2+4=3(7)^2\text{ + 4} \\ =\text{ 151} \\ \\ \text{when x = 8} \\ 3i^2+4=3(8)^2^{}\text{ + 4} \\ =\text{ }196 \\ \\ \text{when x = 9} \\ 3i^2+4=3(9)^2\text{ + 4} \\ =\text{ 247} \end{gathered}$

Next we will add all the results of the term:

$\begin{gathered} \sum ^9_{i=1}(3i^2\text{ + 4)=7 + 16 + 31 + }52\text{ + 79 + 112 + 151 + 196 + 247} \\ \sum ^9_{i=1}(3i^2\text{ + 4)= 89}1\text{ (option }D) \end{gathered}$