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T1101 received 2 Kudos for post Re: (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=?.

GiverPostDate
JS1290Re: (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=?09-Dec-2018
regglestonRe: (1^2+2^2+3^3+…+20^2) / (1+2+3+ …+20)=?24-Oct-2018

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