作业帮an是等差数列,求lim (Sn+Sn+1)/(Sn+Sn-1)lim (Sn+Sn+1)/(Sn+Sn-1)=[n(n+1)/2+(n+1)(n+2)/2]/[n(n+1)/2+n(n-1)/2]=(2n²+4n+2)/2n²=1+2/n+1/n²我就想知道第一步怎么来的
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![an是等差数列,求lim (Sn+Sn+1)/(Sn+Sn-1)lim (Sn+Sn+1)/(Sn+Sn-1)=[n(n+1)/2+(n+1)(n+2)/2]/[n(n+1)/2+n(n-1)/2]=(2n²+4n+2)/2n²=1+2/n+1/n²我就想知道第一步怎么来的](/uploads/image/z/7141629-21-9.jpg?t=an%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E6%B1%82%26%238203%3Blim+%28Sn%2BSn%2B1%29%2F%28Sn%2BSn-1%29lim+%28Sn%2BSn%2B1%29%2F%28Sn%2BSn-1%29%3D%5Bn%28n%2B1%29%2F2%2B%28n%2B1%29%28n%2B2%29%2F2%5D%2F%5Bn%28n%2B1%29%2F2%2Bn%28n-1%29%2F2%5D%3D%282n%26%23178%3B%2B4n%2B2%29%2F2n%26%23178%3B%3D1%2B2%2Fn%2B1%2Fn%26%23178%3B%E6%88%91%E5%B0%B1%E6%83%B3%E7%9F%A5%E9%81%93%E7%AC%AC%E4%B8%80%E6%AD%A5%E6%80%8E%E4%B9%88%E6%9D%A5%E7%9A%84)
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an是等差数列,求lim (Sn+Sn+1)/(Sn+Sn-1)lim (Sn+Sn+1)/(Sn+Sn-1)=[n(n+1)/2+(n+1)(n+2)/2]/[n(n+1)/2+n(n-1)/2]=(2n²+4n+2)/2n²=1+2/n+1/n²我就想知道第一步怎么来的
an是等差数列,求lim (Sn+Sn+1)/(Sn+Sn-1)
lim (Sn+Sn+1)/(Sn+Sn-1)
=[n(n+1)/2+(n+1)(n+2)/2]/[n(n+1)/2+n(n-1)/2]
=(2n²+4n+2)/2n²
=1+2/n+1/n²
我就想知道第一步怎么来的
an是等差数列,求lim (Sn+Sn+1)/(Sn+Sn-1)lim (Sn+Sn+1)/(Sn+Sn-1)=[n(n+1)/2+(n+1)(n+2)/2]/[n(n+1)/2+n(n-1)/2]=(2n²+4n+2)/2n²=1+2/n+1/n²我就想知道第一步怎么来的
an=n
sn=n(n+1)/2