用拆项法求数列(2^2+1)/(2^2-1),(3^2+1)/(3^2-1),(4^2+1)/(4^2-1),…的前n项
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用拆项法求数列(2^2+1)/(2^2-1),(3^2+1)/(3^2-1),(4^2+1)/(4^2-1),…的前n项之和.
![用拆项法求数列(2^2+1)/(2^2-1),(3^2+1)/(3^2-1),(4^2+1)/(4^2-1),…的前n项](/uploads/image/z/15647410-10-0.jpg?t=%E7%94%A8%E6%8B%86%E9%A1%B9%E6%B3%95%E6%B1%82%E6%95%B0%E5%88%97%282%5E2%2B1%29%2F%282%5E2-1%29%2C%283%5E2%2B1%29%2F%283%5E2-1%29%2C%284%5E2%2B1%29%2F%284%5E2-1%29%2C%E2%80%A6%E7%9A%84%E5%89%8Dn%E9%A1%B9)
a(n) = [(n+1)^2 + 1]/[(n+1)^2 - 1] = 1 + 2/[(n+1)^2 - 1]
=1 + 2/[(n+2)n]
=1 + 1/n - 1/(n+2),
s(n) = a(1)+a(2)+a(3)+a(4)+...+a(n-3)+a(n-2)+a(n-1)+a(n)
=n + [1/1-1/3 + 1/2-1/4 + 1/3-1/5 + 1/4-1/6 + ...+ 1/(n-3)-1/(n-1) + 1/(n-2)-1/n + 1/(n-1)-1/(n+1) + 1/n-1/(n+2)]
=n + 1/1 + 1/2 - 1/(n+1) - 1/(n+2)
=n + 3/2 - 1/(n+1) - 1/(n+2)
=1 + 2/[(n+2)n]
=1 + 1/n - 1/(n+2),
s(n) = a(1)+a(2)+a(3)+a(4)+...+a(n-3)+a(n-2)+a(n-1)+a(n)
=n + [1/1-1/3 + 1/2-1/4 + 1/3-1/5 + 1/4-1/6 + ...+ 1/(n-3)-1/(n-1) + 1/(n-2)-1/n + 1/(n-1)-1/(n+1) + 1/n-1/(n+2)]
=n + 1/1 + 1/2 - 1/(n+1) - 1/(n+2)
=n + 3/2 - 1/(n+1) - 1/(n+2)
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