Answer:
Option 4 which is 1,430.
Step-by-step explanation:
Step 1
In this this step we will break up the terms in the summation symbol using the rules for manipulation summations.
[tex]\sum_{i=1}^{n}(aX^2_i+k)=\sum_{i=1}^{n}(aX)+\sum_{i=1}^{n}k\\\\\sum_{i=1}^{n}(aX^2_i+k)=a\sum_{i=1}^{n}X^2+nk.[/tex]
This means that for our case, the sum can be simplified to be,
[tex]\sum_{i=1}^{11}(3i^2_i-8)=3\sum_{i=1}^{11}i^2+11\times(-8)\\\\\sum_{i=1}^{11}(3i^2_i-8)=3\sum_{i=1}^{11}i^2-88[/tex]
Step 2
In this step we use the formula for the sum of the first n squares to evaluate the first part of the sum in step 2. The formula for the sum of the first n squares is,
[tex]\sum_{i=1}^{n}n^2=\frac{n(n+1)(2n+1)}{6} \\\\\implies\sum_{i=1}^{11}i^2=\frac{11(11+1)(2\times 11+1)}{6} =506[/tex]
Step 3
In this step we use the result from step 1 to evaluate the sum.[tex]3\sum_{i=1}^{11}i^2-88=3(506)-88=1,430.[/tex]
The correct answer is Option 4,which is 1,430.
