数列{an}的通项an=n²-n,求sn?_数学解答

数列{an}的通项an=n²-n,求sn?

人气：235 ℃　时间：2020-05-15 07:47:05

解答

a1 = 1² - 1a2 = 2² - 2a3 = 3² - 3...an = n² -nSn = (1²+2²+...+n²) - (1+2+..+n) =n(n+1)(2n+1)/6 - n(n+1)/2 =n(n²-1)/3