Question
Evaluate \(\frac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \frac{1}{2 \cdot 3 \cdot 4 \cdot 5} + \frac{1}{3 \cdot 4 \cdot 5 \cdot 6} + … + \)up to n terms.
Solution:
General term \( T_r = \frac{1}{r(r+1)(r+2)(r+3)} \)
\( = \frac{1}{3} \times \frac{r + 3 – r}{r(r+1)(r+2)(r+3)}\)
\( = \frac{1}{3} [\frac{1}{r(r+1)(r+2)} – \frac{1}{(r+1)(r+2)(r+3)} ]\)
Now \(T_1 = \frac{1}{3} [\frac{1}{1 \cdot 2 \cdot 3} – \frac{1}{2 \cdot 3 \cdot 4}]\)
\(T_2 = \frac{1}{3} [\frac{1}{2 \cdot 3 \cdot 4} – \frac{1}{3 \cdot 4 \cdot 5}]\)
\(T_3 = \frac{1}{3} [\frac{1}{3 \cdot 4 \cdot 5} – \frac{1}{4 \cdot 5 \cdot 6}]\)
\(\cdot\)
\(\cdot\)
\(\cdot\)
\(T_{n-1} = \frac{1}{3} [\frac{1}{(n-1) \cdot (n) \cdot (n+1)} – \frac{1}{n \cdot (n+1) \cdot (n+2)}]\)
\(T_n = \frac{1}{3} [\frac{1}{n \cdot (n+1) \cdot (n+2)} – \frac{1}{(n+1) \cdot (n+2) \cdot (n+3)}]\)
\(T_1 + T_2 + T_3 + … + T_{n-1} + T_n = \sum_{n=1}^{n}T_r\)
Adding them terms will cancel out each others and only first and last terms remain.
\( \sum_{n=1}^{n}T_r = \frac{1}{3}[\frac{1}{1 \cdot 2 \cdot 3} – \frac{1}{(n+1) \cdot (n+2) \cdot (n+3)}] \)