Question Number 219093 by hardmath last updated on 19/Apr/25 Commented by hardmath last updated on 19/Apr/25
Question Number 219025 by universe last updated on 18/Apr/25 Answered by vnm last updated on 19/Apr/25
Question Number 219003 by hardmath last updated on 17/Apr/25 Answered by maths2 last updated on 18/Apr/25 $${S}_{{p}} =\Sigma\Sigma\underset{{k}=\mathrm{0}} {\sum}\frac{{k}^{\mathrm{2}} }{\left({n}−{k}\right)^{\mathrm{2}} +{k}^{\mathrm{2}} }=\Sigma\Sigma\underset{{k}=\mathrm{0}} {\sum}\frac{\left({n}−{k}\right)^{\mathrm{2}} }{\left({n}−{k}\right)^{\mathrm{2}} +{k}^{\mathrm{2}}…
Question Number 218997 by hardmath last updated on 17/Apr/25 Answered by MrGaster last updated on 19/Apr/25
Question Number 218970 by Nicholas666 last updated on 17/Apr/25
Question Number 218854 by Rojarani last updated on 16/Apr/25 Answered by Hamada1969 last updated on 16/Apr/25
Question Number 218848 by hardmath last updated on 16/Apr/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 218855 by Lekhraj last updated on 16/Apr/25
Question Number 218849 by hardmath last updated on 16/Apr/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 218850 by hardmath last updated on 16/Apr/25 Answered by MrGaster last updated on 17/Apr/25 $${T}_{{n}} =\frac{\mathrm{1}}{\mathrm{ln}\left(\underset{{r}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{{r}^{\mathrm{2}} }{\mathrm{1}+{r}^{\mathrm{2}} }\right)}=\frac{\mathrm{1}}{\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{ln}\left(\frac{{r}^{\mathrm{2}} }{\mathrm{1}+{r}^{\mathrm{2}}…