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×× 1+2+3+1²+2²+3²+1³+2³+3³ k=n(n1)/2 k²= n(n1)(2n1)/6 k³={n(n1)/2}² kk=1n 1101+2+3++10 (1+10)×10÷255 71100 5050 1nnn+122 22 1²+2²+3²+=k²n(n+1)(2n+1)/6
18 15 2022.02.26 © nihonbungeisha.All rights reserved.
18 15 2022.02.26 © nihonbungeisha.All rights reserved.
Q A×B×C×D× A× B× C× D× 72350TV×HPMinstrel ©PresidentKUMA/gettyimages©ako maruo/gettyimages
PCRA APCR A 10060 PCRPCR AA A & URLhttps://www.shimada-law.com/
RIAJ60005002 JASRAC6700101198Y45039 © Fuji News Network, Inc. All rights reserved.