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.

$$\sum_{n=1}^{n}T_r = \frac{1}{3}[\frac{1}{1 \cdot 2 \cdot 3} – \frac{1}{(n+1) \cdot (n+2) \cdot (n+3)}]$$