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

QUIZ:  If the sum of the numbers 1 to 100 can  be calculated with Carl Gauss′s method,  then How do we calculate the sum of the  numbers starting from 2 and going up to  100 by twos in an easy way with the same  method?  Or 2+4+6+8+10+...+100=?

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

Commented by Frix last updated on 03/Feb/23

2+4+6+...+100=2×(1+2+3+...+50)  This is too easy

$$\mathrm{2}+\mathrm{4}+\mathrm{6}+...+\mathrm{100}=\mathrm{2}×\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+...+\mathrm{50}\right) \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{too}\:\mathrm{easy} \\ $$

Commented by ajfour last updated on 04/Feb/23

1   +2    +3+..........+24+25  50+49+48+.........+27+26  ______________________  51+51+51+..........+51+51  =25×51=25(50+1)=1250+25

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

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