数列1/1²+2.1/2²+4,1/3²+6,1/4²+8的前n项和等于.还有1
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数列1/1²+2.1/2²+4,1/3²+6,1/4²+8的前n项和等于.还有1/n^2+2n=1/n(n+2)怎么变成1/2(1/n-1/n+2)的.为什么要乘上1/2呢?
![数列1/1²+2.1/2²+4,1/3²+6,1/4²+8的前n项和等于.还有1](/uploads/image/z/19054150-70-0.jpg?t=%E6%95%B0%E5%88%971%EF%BC%8F1%26%23178%3B%2B2.1%EF%BC%8F2%26%23178%3B%2B4%2C1%EF%BC%8F3%26%23178%3B%2B6%2C1%EF%BC%8F4%26%23178%3B%2B8%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E7%AD%89%E4%BA%8E.%E8%BF%98%E6%9C%891)
(1/2)[1/n-1/(n+2)]
=(1/2) { (n+2 -n)/[n(n+1)] }
= 1/[n(n+1)]
an =(1/2)[1/n-1/(n+2)]
Sn = a1+a2+...+an
=(1/2) [ 1+ 1/2 - 1/(n+1) -1/(n+2)]
再问: 1/2)[1/n-1/(n+2)] =(1/2) { (n+2 -n)/[n(n+1)] } = 1/[n(n+1)] 这是什么?我还是不懂1/n^2+2n怎么等于1/2(1/n-1/n+2
再答: (1/2)[1/n-1/(n+2)] =(1/2) { (n+2 -n)/[n(n+2)] } (通分母) = 1/[n(n+2)]
=(1/2) { (n+2 -n)/[n(n+1)] }
= 1/[n(n+1)]
an =(1/2)[1/n-1/(n+2)]
Sn = a1+a2+...+an
=(1/2) [ 1+ 1/2 - 1/(n+1) -1/(n+2)]
再问: 1/2)[1/n-1/(n+2)] =(1/2) { (n+2 -n)/[n(n+1)] } = 1/[n(n+1)] 这是什么?我还是不懂1/n^2+2n怎么等于1/2(1/n-1/n+2
再答: (1/2)[1/n-1/(n+2)] =(1/2) { (n+2 -n)/[n(n+2)] } (通分母) = 1/[n(n+2)]
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