Question Number 186348 by test1234 last updated on 03/Feb/23

$$\mathrm{Q}{UIZ}: \\ $$$${If}\:{the}\:{sum}\:{of}\:{the}\:{numbers}\:\mathrm{1}\:{to}\:\mathrm{100}\:{can} \\ $$$${be}\:{calculated}\:{with}\:{Carl}\:{Gauss}'{s}\:{method}, \\ $$$${then}\:{How}\:{do}\:{we}\:{calculate}\:{the}\:{sum}\:{of}\:{the} \\ $$$${numbers}\:{starting}\:{from}\:\mathrm{2}\:{and}\:{going}\:{up}\:{to} \\ $$$$\mathrm{100}\:{by}\:{twos}\:{in}\:{an}\:{easy}\:{way}\:{with}\:{the}\:{same} \\ $$$${method}? \\ $$$${Or}\:\mathrm{2}+\mathrm{4}+\mathrm{6}+\mathrm{8}+\mathrm{10}+…+\mathrm{100}=? \\ $$
Commented by Frix last updated on 03/Feb/23

$$\mathrm{2}+\mathrm{4}+\mathrm{6}+…+\mathrm{100}=\mathrm{2}×\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+…+\mathrm{50}\right) \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{too}\:\mathrm{easy} \\ $$
Commented by ajfour last updated on 04/Feb/23

$$\mathrm{1}\:\:\:+\mathrm{2}\:\:\:\:+\mathrm{3}+……….+\mathrm{24}+\mathrm{25} \\ $$$$\mathrm{50}+\mathrm{49}+\mathrm{48}+………+\mathrm{27}+\mathrm{26} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\mathrm{51}+\mathrm{51}+\mathrm{51}+……….+\mathrm{51}+\mathrm{51} \\ $$$$=\mathrm{25}×\mathrm{51}=\mathrm{25}\left(\mathrm{50}+\mathrm{1}\right)=\mathrm{1250}+\mathrm{25} \\ $$