已知等差数列{an}中,S₂=16,S₄=24,求数列{│an│}的前n项和An
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已知等差数列{an}中,S₂=16,S₄=24,求数列{│an│}的前n项和An
![已知等差数列{an}中,S₂=16,S₄=24,求数列{│an│}的前n项和An](/uploads/image/z/4362506-26-6.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E4%B8%AD%2CS%26%238322%3B%3D16%2CS%26%238324%3B%3D24%2C%E6%B1%82%E6%95%B0%E5%88%97%EF%BD%9B%E2%94%82an%E2%94%82%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CAn)
等差数列{an}
a(n) = a1 + (n - 1)d
前n项和S(n) = (2*a1 + (n-1)d) * n / 2 ---------记忆方法:梯形公式:(首项+末项)×项数/2
所以:a(2) =a1 + d =16 ①
a(4) =a1 + 3d =24 ②
a1 = 12 d = 4 ---------------其实 d = (am - an) / (m - n) = (24 - 16) / (4 - 2)
a(n) = a1 + (n - 1)d = 12 + 4(n - 1) = 4n + 8
S(n) = (2*a1 + (n-1)d) * n / 2 = (2*12 + 4(n-1)) * n / 2 = n(2n + 10)
a(n) = a1 + (n - 1)d
前n项和S(n) = (2*a1 + (n-1)d) * n / 2 ---------记忆方法:梯形公式:(首项+末项)×项数/2
所以:a(2) =a1 + d =16 ①
a(4) =a1 + 3d =24 ②
a1 = 12 d = 4 ---------------其实 d = (am - an) / (m - n) = (24 - 16) / (4 - 2)
a(n) = a1 + (n - 1)d = 12 + 4(n - 1) = 4n + 8
S(n) = (2*a1 + (n-1)d) * n / 2 = (2*12 + 4(n-1)) * n / 2 = n(2n + 10)
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