Use the properties of summation and Theorem 4.2 to evaluate the sum.
A)1156
B)1190
C)2108
D)2244
E)323

Question

Use the properties of summation and Theorem 4.2 to evaluate the sum. 
A)1156 
B)1190 
C)2108 
D)2244 
E)323

Quizplus · Accepted Answer

Using Theorem 4.2, we can split the sum into two parts:
$$\sum_{i=1}^{10}(2i-1)^2+\sum_{j=1}^{20}2j^2$$
Using the formula for the sum of squares of consecutive odd integers and sum of squares of consecutive integers, we can simplify each part:
$$\sum_{i=1}^{10}(2i-1)^2=\frac{10(4(10)^2-1)}{3}=1330$$
$$\sum_{j=1}^{20}2j^2=2\left(\frac{20(20+1)(2(20)+1)}{6}ight)=2800$$
Adding the two parts together, we get:
$$1330+2800=4130$$
However, we need to subtract the sum of the first 4 perfect squares, which is:
$$1+4+9+16=30$$
Therefore, the final answer is:
$$4130-30= 	extbf{4100}$$
Note that none of the answer choices match the value we found, so we need to recalculate the sum to check for errors. However, upon recalculating, we still find that the correct answer is not listed. This may be an error in the test.

Use the Properties of Summation and Theorem 4