Question Number 192550 by mokys last updated on 20/May/23 | ||
![]() | ||
$${how}\:{can}\:{find}\:{the}\:{sum}\:\underset{{i}=\mathrm{1}} {\overset{{r}} {\sum}}\left(\mathrm{2}{v}_{{i}} +\mathrm{1}\right)\:? \\ $$ | ||
Answered by a.lgnaoui last updated on 20/May/23 | ||
![]() | ||
$$\mathrm{S}_{\mathrm{i}} =\sum_{\mathrm{i}=\mathrm{1}} ^{\mathrm{i}=\mathrm{r}} \left(\mathrm{2v}_{\mathrm{i}} +\mathrm{1}\right)=\mathrm{1}+\mathrm{2}\sum_{\mathrm{i}=\mathrm{1}} ^{\mathrm{i}=\mathrm{r}} \mathrm{v}_{\mathrm{i}} \\ $$$$=\mathrm{1}+\mathrm{2}\left(\mathrm{v}_{\mathrm{1}} +\mathrm{v}_{\mathrm{2}} +\mathrm{v}_{\mathrm{3}} +....\mathrm{v}_{\mathrm{r}} \right) \\ $$ | ||
Commented by JDamian last updated on 20/May/23 | ||
![]() | ||
$${wrong} \\ $$$$=\boldsymbol{{r}}+\mathrm{2}\left(\mathrm{v}_{\mathrm{1}} +\mathrm{v}_{\mathrm{2}} +\mathrm{v}_{\mathrm{3}} +....\mathrm{v}_{\mathrm{r}} \right) \\ $$ | ||
Commented by a.lgnaoui last updated on 20/May/23 | ||
![]() | ||
$$\mathrm{2}\sum_{\mathrm{i}=\mathrm{1}} ^{\mathrm{i}=\mathrm{r}} \mathrm{v}_{\mathrm{i}} +\left(\mathrm{1}+\mathrm{1}+\mathrm{1}+...+\mathrm{1}\right)\:\:\:\:\:\:\:\left(\mathrm{1}×\mathrm{r}\right)\: \\ $$$$=\mathrm{r}+\mathrm{2}\sum_{\mathrm{i}=\mathrm{1}} ^{\mathrm{i}=\mathrm{r}} \mathrm{v}_{\mathrm{i}} \\ $$$$\mathrm{thanks}\:\:\:\mathrm{for}\:\mathrm{help} \\ $$ | ||