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
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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