来自胡世光的问题
(1)求和:Sn=112+214+318+…+(n+12n).(2)an=1n(n+2),n∈N+,求此数列的前n项和Sn.
(1)求和:Sn=112+214+318+…+(n+12n).
(2)an=1n(n+2),n∈N+,求此数列的前n项和Sn.
1回答
2020-11-1712:16
(1)求和:Sn=112+214+318+…+(n+12n).(2)an=1n(n+2),n∈N+,求此数列的前n项和Sn.
(1)求和:Sn=112+214+318+…+(n+12n).
(2)an=1n(n+2),n∈N+,求此数列的前n项和Sn.
(1)Sn=(1+2+…+n)+(12+122+…+12n)=n(n+1)2+12(1-12n)1-12=n(n+1)2+1-12n.(2)an=12(1n-1n+2),∴此数列的前n项和Sn=12[(1-13)+(12-14)+(13-15)+…+(1n-1-1n+1)+(1n-1n+2)]=12(1+12-1n+1-1n+2)=34-2n+32(n+1)(...