evaluate the series 4 sigma n=1 n+4
26
10
16
-6
The series can be written as:
4 * (1+4) + 4 * (2+4) + 4 * (3+4) + ...
Expanding each term gives:
4 + 16 + 28 + ...
To find the sum of this series, we see that each term is increasing by 12. Therefore, the sum can be calculated as follows:
4 + 16 + 28 + ... + 4n = 4n + 12 * (1+2+3+...+n-1)
Using the formula for the sum of an arithmetic series:
4n + 12 * ((n-1)n)/2 = 4n + 6n(n-1) = 10n^2 - 2n
Therefore, when n = 1, the sum is 10(1)^2 - 2(1) = 10 - 2 = 8
when n = 2, the sum is 10(2)^2 - 2(2) = 40 - 4 = 36
when n = 3, the sum is 10(3)^2 - 2(3) = 90 - 6 = 84
when n = 4, the sum is 10(4)^2 - 2(4) = 160 - 8 = 152
So the evaluations of the series are:
n=1: 4 sigma n=1 n+4 = 8
n=2: 4 sigma n=1 n+4 = 36
n=3: 4 sigma n=1 n+4 = 84
n=4: 4 sigma n=1 n+4 = 152