# If $${\sum}_{r=1}^{n‎} T_r$$ = $$n\over 8$$ (n + 1)(n + 2)(n + 3), then find $${\sum}_{r=1}^{n‎}$$$$1\over T_r$$

## Solution :

$$\because$$ $$T_n$$ = $$S_n – S_{n-1}$$

= $${\sum}_{r=1}^{n‎} T_r$$ – $${\sum}_{r=1}^{n‎ – 1} T_r$$

= $$n(n+1)(n+2)(n+3)\over 8$$ – $$(n-1)(n)(n+1)(n+2)\over 8$$

= $$n(n+1)(n+2)\over 8$$[(n+3) – (n-1)] = $$n(n+1)(n+2)\over 8$$(4)

$$T_n$$ = $$n(n+1)(n+2)\over 2$$

$$\implies$$ $$1\over T_n$$ = $$2\over n(n+1)(n+2)$$ = $$(n+2)-n\over n(n+1)(n+2)$$

= $$1\over n(n+1)$$ – $$1\over (n+1)(n+2)$$

Let $$V_n$$ = $$1\over n(n+1)$$

$$\therefore$$ $$1\over T_n$$ = $$V_n$$ – $$V_{n+1}$$

Putting n = 1, 2, 3, …… n

$$\implies$$ $$1\over T_1$$ + $$1\over T_2$$ + $$1\over T_3$$ + ….. + $$1\over T_n$$ = $$V_1$$ – $$V_{n+1}$$

= $${\sum}_{r=1}^{n‎}$$$$1\over T_r$$ = $$n^2 + 3n\over 2(n+1)(n+2)$$

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