Find the sum.
$$\sum _ { k = 1 } ^ { 3 } \frac { 1 } { k ^ { 2 } + 5 }$$

Question

Quizplus · Accepted Answer

The sum can be calculated by plugging in the values of $k$ from 1 to 3 into the given formula and adding the results:$$\sum _ { k = 1 } ^ { 3 } \frac { 1 } { k ^ { 2 } + 5 } = \frac { 1 } { 1 ^ { 2 } + 5 } + \frac { 1 } { 2 ^ { 2 } + 5 } + \frac { 1 } { 3 ^ { 2 } + 5 } = \frac { 1 } { 6 } + \frac { 1 } { 9 } + \frac { 1 } { 14 } = \frac { 9 + 6 + 4 } { 63 } = \frac { 19 } { 63 }.$$However, the correct calculation should be:$$\frac { 1 } { 6 } + \frac { 1 } { 9 } + \frac { 1 } { 14 } = \frac { 14 + 9 + 6 } { 126 } = \frac { 29 } { 126 }.$$But this calculation is incorrect. The correct calculation is:$$\frac { 1 } { 1^2 + 5 } + \frac { 1 } { 2^2 + 5 } + \frac { 1 } { 3^2 + 5 } = \frac { 1 } { 6 } + \frac { 1 } { 9 } + \frac { 1 } { 14 } = \frac { 14 + 18 + 9 } { 126 } = \frac { 41 } { 126 }.$$The correct sum is actually:$$\frac { 1 } { 6 } + \frac { 1 } { 9 } + \frac { 1 } { 14 } = \frac { 14 + 18 + 9 } { 126 } = \frac { 41 } { 126 } = \frac { 22 } { 63 }.$$So, the correct answer is $\frac { 22 } { 63 }$, which corresponds to option A.

Find the Sum ∑k=131k2+5\sum _ { k = 1 } ^ { 3 } \frac { 1 } { k ^ { 2 } + 5 }k=1∑3​k2+51​

Find the Sum $\sum _ { k = 1 } ^ { 3 } \frac { 1 } { k ^ { 2 } + 5 }$