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0:-1,r=e.isIntersecting?e.intersectionRatio
0:-1;if(n!==r)for(var i=0;i0&&function(t,e,n,r){var i=document.getElementsByClassName(t);if(i.length>0)for(var o=0;o展开全部
解析分式的上下同时x1/n化为(1+2/n)/(2+1/n)x趋于多少如果是无穷则为1/2希望对你有帮助学习进步O(∩_∩)O谢谢
展开全部limn→∞(n+2)/(2n+1)=limn→∞【1/2+3/2（2n+1）】=limn→∞1/2
展开全部n趋向于无穷大时，极限是二分之一
收起
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an=(2n-1)*(1/2)^(n-3)典型的等差乘等比，错位相减法：Sn=1*(1/2)^(-2)+3*(1/2)^(-1)+5*(1/2)^0+。。。+(2n-3)*(1/2)^(n-4)+(2n-1)*(1/2)^(n-3)(1/2)Sn=1*(1/2)^(-1)+3*(1/2)^0+。。。+(2n-5)*(1/2)^(n-4)+(2n-3)*(1/2)^(n-3)+(2n-1)2^(2-n)两式相减：(1/2)Sn=1*(1/2)^(-2)+[2*(1/2)^(-1)+2*(1/2)^0+...+2*(1/2)^(n-3)]-(2n-1)*2^(2-n)=4+4[1-(1/2)^(n-1)]/[1-(1/2)]-(2n-1)*2^(2-n)=4+8-8*2^(1-n)-(2n-1)*2^(2-n)=4+8-4*2^(2-n)-(2n-1)*2^(2-n)=12-(2n+3)*2^(2-n)所以：Sn=24-(2n+3)*2^(3-n)希望能帮到你，如果不懂，请Hi我，祝学习进步！
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收起Sn=(1×2^1+3×2^2+5×2^3+……+(2n-3)×2^(n-1)+(2n-1)×2^n
)/8
(1)两边同时自乘以公比2得2Sn=
(
1×2^2+3×2^3+5×2^4+……+(2n-3)×2^n+(2n-1)×2^(n+1)
)
/8
(2)(1)式减去（2）式得－8Sn=1×2^1+[2×2^2+2×2^3+……2×2^(n-1)+(2n-1)×2^n]－(2n-1)×2^(n+1)
=[2×2^1+2×2^2+2×2^3+……2×2^(n-1)+(2n-1)×2^n]－(2n-1)×2^(n+1)－1×2^1=4(1-2^n)/(1-2)－(2n-1)×2^(n+1)－1×2^1=－4+2×2^(n+1)－(2n-1)×2^(n+1)－2=－6－（2n－3）×2^(n+1)Sn=(6+（2n－3）×2^(n+1))/8
第n项的1/2倍=(2n-1)/2^(n-2)第n+1项=(2n+1)/2^(n-2)于是a n+1-1/2a n=2^(3-n)a2-1/2a1=2^2a3-1/2a2=2^1......a n+1-1/2a n=2^(3-n)上式累加得到(a2+a3+...+a n+1)-1/2(a1+a2+...+an)=-1/2a1+1/2a2+1/2a3+...+1/2an+a n+1
=4(1-2^n)/(1-2)=4(2^n-1)由于a1=4,上式左边加上a1就得到1/2Sn+a n+1=2^(n+2)从而Sn+2a n+1=2^(n+3)又因为2a n+1=(2n+1)/2^(n-3)所以Sn=2^(n+3)-(2n+1)/2^(n-3)
第n项的1/2倍=(2n-1)/2^(n-2)第n+1项=(2n+1)/2^(n-2)于是a n+1-1/2a n=2^(3-n)a2-1/2a1=2^2a3-1/2a2=2^1......a n+1-1/2a n=2^(3-n)上式累加得到(a2+a3+...+a n+1)-1/2(a1+a2+...+an)=-1/2a1+1/2a2+1/2a3+...+1/2an+a n+1
=4(1-2^n)/(1-2)=4(2^n-1)由于a1=4,上式左边加上a1就得到1/2Sn+a n+1=2^(n+2)从而Sn+2a n+1=2^(n+3)又因为2a n+1=(2n+1)/2^(n-3)所以Sn=2^(n+3)-(2n+1)/2^(n-3) 这就对着里！