高中数学必修5数列题目急求 在线等 谢谢!!!
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8. An +3= 2( An-1 +3), A1 +3=4,An +3是首项为4,公比为2的等比数列
An +3 =4*2^(n-1)=2^(n+1), An=2^(n+1)-3
Sn=4(1-2^n)/(1-2)-3n=2^(n+2)-3n-4
9.原式=[(第一项+第89项)+(2项+88项)+……+(44项+46项)]+45项
=1*44+1/2=89/2
10.原式=1+(-2^2+3^2)+(-4^2+5^2)+……+(-98^2+99^2)-100^2
=1+5+9+13+……+197-100^2=(1+197)[(197-1)/4+1]/2-100^2=-5050
11.an=1+2/[n(n+2)]
Sn=n+2[1/(1*3)+1/(2*4)+……+1/ n(n+2) ]
=n+[1/1-1/3+1/2-1/4+1/3-1/5+…+1/(n-3)-1/(n-1)+1/(n-2)-1/n+1/(n-1)-1/(n+1)+1/n-1/(n+2)]
=n+[1+1/2-1/(n+1)-1/(n+2)]
=n+3/2-1/(n+1)-1/(n+2)
An +3 =4*2^(n-1)=2^(n+1), An=2^(n+1)-3
Sn=4(1-2^n)/(1-2)-3n=2^(n+2)-3n-4
9.原式=[(第一项+第89项)+(2项+88项)+……+(44项+46项)]+45项
=1*44+1/2=89/2
10.原式=1+(-2^2+3^2)+(-4^2+5^2)+……+(-98^2+99^2)-100^2
=1+5+9+13+……+197-100^2=(1+197)[(197-1)/4+1]/2-100^2=-5050
11.an=1+2/[n(n+2)]
Sn=n+2[1/(1*3)+1/(2*4)+……+1/ n(n+2) ]
=n+[1/1-1/3+1/2-1/4+1/3-1/5+…+1/(n-3)-1/(n-1)+1/(n-2)-1/n+1/(n-1)-1/(n+1)+1/n-1/(n+2)]
=n+[1+1/2-1/(n+1)-1/(n+2)]
=n+3/2-1/(n+1)-1/(n+2)
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