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Question Number 192367 by Rupesh123 last updated on 15/May/23

Answered by mehdee42 last updated on 15/May/23

$$t_r=r-\frac{3}{2}\Rightarrow t_{r+1}=r-\frac{1}{2}\mathrm{\&}{t}_{r+2}=r+\frac{1}{2}\mathrm{\&}{t}_{r+3}=r+\frac{3}{2}$$$$\Rightarrow t_r{t}_{r+1}{t}_{r+2}{t}_{r+3}=r^4-\frac{5}{2}{r}^2+\frac{25}{26}$$$$\Rightarrow \underset{r=1}{\sum ^n}\sqrt{1+t_r{t}_{r+1}{t}_{r+2}{t}_{r+3}}=\underset{r=1}{\sum ^n}\sqrt{r^4-\frac{5}{2}{r}^2+\frac{25}{16}}$$$$=\underset{r=1}{\sum ^n}(r^2-\frac{5}{4})=\frac{n(n+1)(2n+1)}{6}-\frac{5}{4}n✓$$

Commented by Rupesh123 last updated on 15/May/23

Nice, sir!

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