Evaluate the series
-$\sum_{\mathrm{i}=3}^{6} \frac{\left(\mathrm{i}^{2}-2\right)}{2}$
A) 39
B) 40
C) 78
D) 21

Question

Quizplus · Accepted Answer

The series can be evaluated by substituting the values of i from 3 to 6 into the given formula and summing the results:For i=3: $\frac{(3^2-2)}{2} = \frac{7}{2}$For i=4: $\frac{(4^2-2)}{2} = \frac{14}{2} = 7$For i=5: $\frac{(5^2-2)}{2} = \frac{23}{2}$For i=6: $\frac{(6^2-2)}{2} = \frac{34}{2} = 17$Summing these results gives: $\frac{7}{2} + 7 + \frac{23}{2} + 17 = \frac{14 + 46 + 34}{2} = \frac{94}{2} = 47$However, the correct calculation should be:For i=3: $\frac{(3^2-2)}{2} = \frac{7}{2}$For i=4: $\frac{(4^2-2)}{2} = \frac{14}{2} = 7$For i=5: $\frac{(5^2-2)}{2} = \frac{23}{2}$For i=6: $\frac{(6^2-2)}{2} = \frac{34}{2} = 17$Summing these results gives: $\frac{7}{2} + 7 + \frac{23}{2} + 17 = 7 + 7 + 11.5 + 17 = 42.5$It seems there was a mistake in my initial calculation. Let's correctly sum the values:For i=3: $\frac{(3^2-2)}{2} = \frac{9-2}{2} = \frac{7}{2} = 3.5$For i=4: $\frac{(4^2-2)}{2} = \frac{16-2}{2} = 7$For i=5: $\frac{(5^2-2)}{2} = \frac{25-2}{2} = 11.5$For i=6: $\frac{(6^2-2)}{2} = \frac{36-2}{2} = 17$Summing these results gives: $3.5 + 7 + 11.5 + 17 = 39$Therefore, the correct sum of the series is 39.

Evaluate the Series - ∑i=36(i2−2)2\sum_{\mathrm{i}=3}^{6} \frac{\left(\mathrm{i}^{2}-2\right)}{2}∑i=36​2(i2−2)​ A) 39 B) 40 C) 78 D) 21

Evaluate the Series
- $\sum_{\mathrm{i}=3}^{6} \frac{\left(\mathrm{i}^{2}-2\right)}{2}$ A) 39
B) 40
C) 78
D) 21