Find the first 5 partial sums of the series ∑n=2 to ∞ 1/(n(n+4))?
I got 1/12 for the first one but when I put 1/21 for the next one it says it's wrong please help
1/21 is just the 2nd term in the sequence, but is not a partial sum
S1 = 1/12
S2 = 1/12 + 1/21 = 11/84
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To find the partial sums of the series ∑n=2 to ∞ 1/(n(n+4)), you can follow these steps:
1. Start by calculating the first term, which corresponds to n = 2. Plug in n = 2 into the series expression:
Sum_2 = 1 / (2(2+4)) = 1/12
2. To find the second term, add the second term in the series: n = 3.
Sum_2 + Sum_3 = 1/12 + 1 / (3(3+4))
3. Simplify the expression: 1 / (3(3+4)) = 1/21
So, the second partial sum is 1/12 + 1/21.
Now, let's compute the value of the second sum:
Sum_2 + Sum_3 = 1/12 + 1/21.
To add these fractions, we need to find a common denominator, which in this case is 84.
Converting the fractions to have a common denominator:
(7/84) + (4/84) = 11/84.
Therefore, the second partial sum is 11/84.
Following the same logic, the third partial sum would be:
Sum_2 + Sum_3 + Sum_4 = 1/12 + 1/21 + 1 / (4(4+4)) = 1/84.
Sum_2 + Sum_3 + Sum_4 + Sum_5 = 1/12 + 1/21 + 1/84 + 1 / (5(5+4)) = 23/420.
Hence, the first five partial sums of the series ∑n=2 to ∞ 1/(n(n+4)) are:
Sum_2 = 1/12
Sum_2 + Sum_3 = 11/84
Sum_2 + Sum_3 + Sum_4 = 1/84
Sum_2 + Sum_3 + Sum_4 + Sum_5 = 23/420