Studdy Solution
STEP 1
1. The series is an infinite series starting from k=2.
2. The series involves a rational function that can be simplified using partial fraction decomposition.
3. We will need to find a pattern or telescoping nature in the series to evaluate it.
STEP 2
1. Perform partial fraction decomposition on the general term.
2. Simplify the series using the decomposed terms.
3. Identify the telescoping pattern and evaluate the sum.
STEP 3
Perform partial fraction decomposition on the term (3k+1)(3k+4)1.
Assume:
(3k+1)(3k+4)1=3k+1A+3k+4B
Multiply through by the denominator:
1=A(3k+4)+B(3k+1)
Expand and collect like terms:
1=(3A+3B)k+(4A+B)
Set up the system of equations:
3A+3B=0
4A+B=1
Solve for A and B.
STEP 4
From the first equation, 3A+3B=0, we have:
A+B=0
B=−A
Substitute B=−A into the second equation:
4A−A=1
3A=1
A=31
Then, B=−A=−31.
Thus, the partial fraction decomposition is:
(3k+1)(3k+4)1=3k+11/3−3k+41/3
STEP 5
Rewrite the series using the decomposed terms:
k=2∑∞(3k+11/3−3k+41/3)
Factor out the constant 31:
31k=2∑∞(3k+11−3k+41)
STEP 6
Observe the telescoping nature of the series:
The terms will cancel in a telescoping manner, leaving only a few terms from the beginning and the end.
Write out the first few terms to see the pattern:
31((71−101)+(101−131)+(131−161)+⋯)
Notice that all intermediate terms cancel, leaving:
31(71−n→∞lim3n+41)
As n→∞, 3n+41→0.
Therefore, the sum is:
31×71=211
The value of the series is:
211
