Math  /  Calculus

Question10. Calculate the value of the following series: k=21(3k+1)(3k+4)\sum_{k=2}^{\infty} \frac{1}{(3 k+1)(3 k+4)}
ANS:

Studdy Solution

STEP 1

1. The series is an infinite series starting from k=2 k = 2 .
2. The series involves a rational function that can potentially be simplified using partial fraction decomposition.

STEP 2

1. Perform partial fraction decomposition on the general term.
2. Simplify the series using the decomposed form.
3. Evaluate the series to find its sum.

STEP 3

Perform partial fraction decomposition on the general term 1(3k+1)(3k+4)\frac{1}{(3k+1)(3k+4)}. Assume:
1(3k+1)(3k+4)=A3k+1+B3k+4\frac{1}{(3k+1)(3k+4)} = \frac{A}{3k+1} + \frac{B}{3k+4}
Multiply through by the common denominator (3k+1)(3k+4)(3k+1)(3k+4) to clear the fractions:
1=A(3k+4)+B(3k+1)1 = A(3k+4) + B(3k+1)
Expand and collect like terms:
1=(3A+3B)k+(4A+B)1 = (3A + 3B)k + (4A + B)
Equate coefficients to solve for A A and B B :
3A+3B=0(coefficient of k)3A + 3B = 0 \quad \text{(coefficient of } k\text{)} 4A+B=1(constant term)4A + B = 1 \quad \text{(constant term)}

STEP 4

Solve the system of equations:
From 3A+3B=0 3A + 3B = 0 , we have:
A+B=0B=AA + B = 0 \quad \Rightarrow \quad B = -A
Substitute B=A B = -A into 4A+B=1 4A + B = 1 :
4AA=14A - A = 1 3A=13A = 1 A=13A = \frac{1}{3}
Then, B=A=13 B = -A = -\frac{1}{3} .
Thus, the partial fraction decomposition is:
1(3k+1)(3k+4)=1/33k+11/33k+4\frac{1}{(3k+1)(3k+4)} = \frac{1/3}{3k+1} - \frac{1/3}{3k+4}

STEP 5

Rewrite the series using the partial fractions:
k=2(1/33k+11/33k+4)\sum_{k=2}^{\infty} \left( \frac{1/3}{3k+1} - \frac{1/3}{3k+4} \right)
Recognize this as a telescoping series. Write out the first few terms to see the cancellation:
1/371/310+1/3101/313+1/3131/316+\frac{1/3}{7} - \frac{1/3}{10} + \frac{1/3}{10} - \frac{1/3}{13} + \frac{1/3}{13} - \frac{1/3}{16} + \cdots

STEP 6

Observe the cancellation pattern in the telescoping series. The series simplifies to:
k=2(1/33k+11/33k+4)=1/37limn1/33n+4\sum_{k=2}^{\infty} \left( \frac{1/3}{3k+1} - \frac{1/3}{3k+4} \right) = \frac{1/3}{7} - \lim_{n \to \infty} \frac{1/3}{3n+4}
Since the term 1/33n+4\frac{1/3}{3n+4} approaches zero as n n \to \infty , the sum of the series is:
121\frac{1}{21}
The value of the series is:
121 \boxed{\frac{1}{21}}

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