find the value of 1/6 + 1/12 + 1/20 +...+ 1/210 + 1/240

asked by guest
on Oct 22, 2024 at 8:49 am



You asked:

Find the sum of the sequence: \(\frac{1}{6}\), \(\frac{1}{12}\), \(\frac{1}{20}\), \(\ldots\), \(\frac{1}{210}\), \(\frac{1}{240}\)

MathBot Answer:

The sum of the sequence is \(\displaystyle \frac{7}{16}\)


The explicit formula for the nth term of this sequence is \[a_n = \frac{1}{n^{2} + 3 n + 2}\]Summation Formula: $$\begin{aligned}S_n&=\sum_{i=1}^{n} a_{i} \\ &=\sum_{i=1}^{n} \frac{1}{i^{2} + 3 i + 2} \\ &=\frac{n}{2 \left(n + 2\right)} \end{aligned}$$ where $n$ is the is the term number.
Note: Replace all $n$ in $a_n$ in the summation with $i$.

Use the explicit formula to find the index of the final term in the sequence.$$\begin{aligned}\frac{1}{240} &= \frac{1}{n^{2} + 3 n + 2} \\ n &= 14\end{aligned}$$Then use the summation formula to solve.$$\begin{aligned}S_n &= \frac{n}{2 \left(n + 2\right)} \\S_{14} &= \frac{14 \left(2 + 14\right)^{-1}}{2}\\S_{14} &= \frac{7}{16}\end{aligned}$$