Sn=a1+a2+a3+~~~+an=1/2^1+2/2^2+~~~+n/2^n
1/2×Sn=1/2^2+2/2^3+~~~+(n-1)2^n+n/2^(n+1)
Sn-1/2Sn=1/2Sn=1/2+1/2^2+1/2^3+~~~+1/2^n-n/2^(n+1)=(1/2-1/2^n×1/2)/(1-1/2)-n/2^(n+1)
=1-1/2^n-n/2^(n+1)=1-(2+n)/2×1/2^n
∴Sn=2-(2+n)×1/2^n
数列2的n次方分之n的前n项的和
Sn-1\/2Sn=1\/2Sn=1\/2+1\/2^2+1\/2^3+~~~+1\/2^n-n\/2^(n+1)=(1\/2-1\/2^n×1\/2)\/(1-1\/2)-n\/2^(n+1)=1-1\/2^n-n\/2^(n+1)=1-(2+n)\/2×1\/2^n ∴Sn=2-(2+n)×1\/2^n
数列2的n次方分之n的前n项的和
Sn-1\/2Sn=1\/2Sn=1\/2+1\/2^2+1\/2^3+~~~+1\/2^n-n\/2^(n+1)=(1\/2-1\/2^n×1\/2)\/(1-1\/2)-n\/2^(n+1)=1-1\/2^n-n\/2^(n+1)=1-(2+n)\/2×1\/2^n ∴Sn=2-(2+n)×1\/2^n
数学...数列 2的n次方分之2n 其前n项和怎么算?
求和项记作A A乘2约去分母一个2 该式记作B A与B分母相同的对应项约去 可得 希望采纳 祝你学习进步
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