Question Number 202551 by aba last updated on 29/Dec/23
$$\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{i}} =\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{1}} +\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{2}} +…\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{n}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{x}_{\mathrm{1}} +\mathrm{y}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} +\mathrm{y}_{\mathrm{2}} +…+\mathrm{x}_{\mathrm{n}} +\mathrm{y}_{\mathrm{n}} \: \\ $$$$\mathrm{please}\:\mathrm{it}'\mathrm{s}\:\mathrm{correct}\:? \\ $$$$ \\ $$
Commented by mr W last updated on 29/Dec/23
$${it}\:{depends}\:{on}\:{what}\:{you}\:{mean}\:{with} \\ $$$$\left({x}+{y}\right)_{{i}} .\:{if}\:{you}\:{mean}\:\left({x}+{y}\right)_{{i}} ={x}_{{i}} +{y}_{{i}} , \\ $$$${then}\:{it}'{s}\:{certainly}\:{correct}. \\ $$