求解,给好评,感谢
1个回答
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(-1)ⁿ⁻¹·n/2ⁿ=-n·(-½)ⁿ
Sn=1·(-½)+2·(-½)²+...+n·(-½)ⁿ
(-½)Sn=1·(-½)²+2·(-½)³...+(n-1)·(-½)ⁿ+n·(-½)ⁿ⁺¹
Sn-(-½)Sn=(3/2)Sn=(-½)+(-½)²+...+(-½)ⁿ-n·(-½)ⁿ⁺¹
=(-½)[1-(-½)ⁿ]/[1-(-½)] -n·(-½)ⁿ⁺¹
=[(3n+2)/6](-½)ⁿ-⅓
Sn=[(3n+2)/9](-½)ⁿ- (2/9)
∞
∑ (-1)ⁿ⁻¹·n/2ⁿ
n=1
=lim (2/9)-[(3n+2)/[9·(-2)ⁿ]
n→∞
=2/9 -0
=2/9
Sn=1·(-½)+2·(-½)²+...+n·(-½)ⁿ
(-½)Sn=1·(-½)²+2·(-½)³...+(n-1)·(-½)ⁿ+n·(-½)ⁿ⁺¹
Sn-(-½)Sn=(3/2)Sn=(-½)+(-½)²+...+(-½)ⁿ-n·(-½)ⁿ⁺¹
=(-½)[1-(-½)ⁿ]/[1-(-½)] -n·(-½)ⁿ⁺¹
=[(3n+2)/6](-½)ⁿ-⅓
Sn=[(3n+2)/9](-½)ⁿ- (2/9)
∞
∑ (-1)ⁿ⁻¹·n/2ⁿ
n=1
=lim (2/9)-[(3n+2)/[9·(-2)ⁿ]
n→∞
=2/9 -0
=2/9
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