Find the sum of a telescoping series..
4)
_
Q
n=1
7
(4n - 1)(4n + 3)
To find the sum of the telescoping series
(4n - 1)(4n + 3)
we can use partial fraction decomposition to simplify it.
First, we can expand the expression:
(4n - 1)(4n + 3) = 16n^2 + 12n - 4n - 3 = 16n^2 + 8n - 3
Next, we can split the expression using partial fraction decomposition:
(16n^2 + 8n - 3) = A/(4n - 1) + B/(4n + 3)
Multiplying both sides by (4n - 1)(4n + 3), we get:
16n^2 + 8n - 3 = A(4n + 3) + B(4n - 1)
Expanding and collecting like terms:
16n^2 + 8n - 3 = (4A + 4B)n + (3A - B)
Comparing the coefficients of n, we have:
4A + 4B = 16
3A - B = -3
Solving this system of equations, we find A = 1/2 and B = 7/2.
Therefore, we can write the series as:
(4n - 1)(4n + 3) = (1/2)/(4n - 1) + (7/2)/(4n + 3)
Now, we can find the sum of the telescoping series by telescoping:
S = ∑[(1/2)/(4n - 1) + (7/2)/(4n + 3)]
= ∑[(1/2)/(4n - 1)] + ∑[(7/2)/(4n + 3)]
The first summation on the right-hand side can be expressed as:
∑[(1/2)/(4n - 1)] = (1/2) * ∑[1/(4n - 1)]
To calculate this sum, we can use the fact that:
∑[1/(4n - 1)] = ln|4n - 1|/4
Thus, the first term simplifies to:
(1/2) * ∑[1/(4n - 1)] = (1/2) * ln|4n - 1|/4
The second summation on the right-hand side can be expressed as:
∑[(7/2)/(4n + 3)] = (7/2) * ∑[1/(4n + 3)]
To calculate this sum, we can again use the fact that:
∑[1/(4n + 3)] = ln|4n + 3|/4
So, the second term simplifies to:
(7/2) * ∑[1/(4n + 3)] = (7/2) * ln|4n + 3|/4
Now, we can write the sum of the telescoping series:
S = (1/2) * ln|4n - 1|/4 + (7/2) * ln|4n + 3|/4
To find the sum, it would be helpful to know the upper limit of the sum, denoted as N. Once we have the value of N, we can plug it into the expression above and simplify further.