Question Number 192367 by Rupesh123 last updated on 15/May/23
Answered by mehdee42 last updated on 15/May/23
$${t}_{{r}} ={r}−\frac{\mathrm{3}}{\mathrm{2}}\Rightarrow{t}_{{r}+\mathrm{1}} ={r}−\frac{\mathrm{1}}{\mathrm{2}}\:\&\:{t}_{{r}+\mathrm{2}} ={r}+\frac{\mathrm{1}}{\mathrm{2}}\&\:{t}_{{r}+\mathrm{3}} ={r}+\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{t}_{{r}} {t}_{{r}+\mathrm{1}} {t}_{{r}+\mathrm{2}} {t}_{{r}+\mathrm{3}} ={r}^{\mathrm{4}} −\frac{\mathrm{5}}{\mathrm{2}}{r}^{\mathrm{2}} +\frac{\mathrm{25}}{\mathrm{26}} \\ $$$$\Rightarrow\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\sqrt{\mathrm{1}+{t}_{{r}} {t}_{{r}+\mathrm{1}} {t}_{{r}+\mathrm{2}} {t}_{{r}+\mathrm{3}} }=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\sqrt{{r}^{\mathrm{4}} −\frac{\mathrm{5}}{\mathrm{2}}{r}^{\mathrm{2}} +\frac{\mathrm{25}}{\mathrm{16}}} \\ $$$$=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\:\left({r}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}\right)=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}−\frac{\mathrm{5}}{\mathrm{4}}{n}\:\:\checkmark \\ $$
Commented by Rupesh123 last updated on 15/May/23
Nice, sir!