数列求和

Question

如图

Answer 1

Sn＝1²/2+2²/2^2+3²/2^3+……+n²/2^n
∴Sn/2＝1/2^2+2²/2^3+3²/2^4+……+(n-2)²/2^n+n²/2^(n+1)
两式相减，右边错位相减得：
Sn/2＝1/2+ 3/2^2+5/2^3+7/2^4+……+(2n-1)/2^n-n²/2^(n+1)
∴Sn/4＝1/4+ 3/2^3+5/2^4+7/2^5+……+(2n-3)/2^n+(2n-1)/2^(n+1)-n²/2^(n+2)
以上两式再错位相减一次：
Sn/4＝1/2+2/4+2/2^3+2/2^4+2/2^5+……+2/2^n- (n²-2n+1)/2^(n+1)-n²/2^(n+2)
＝1/2+[1/2+1/2^2+1/2^3+1/2^4+……+1/2^(n-1) - (3n²-4n+2)/2^(n+2)
＝3/2-1/2^(n-1) - (3n²-4n+2)/2^(n+2)
∴Sn＝6-4/2^(n-1) - (3n²-4n+2)/2^n ＝6-8/2^n-(3n²-4n+2)/2^n＝6-(3n²-4n+10)/2^n
当n无穷大时，Sn极限为6

Answer 2

公式:首项加末项的和乘项数除以二

Answer 3

我回答对了给我加分啊！

Answer 4

麻烦楼主给数列，谢谢....