已知an=(2n-1)/2^(n-1),求an的前n项和
1个回答
1.
Sn=A1+A2+……+An
=1/2^0+3/2^1+5/2^2……+(2n-1)/2^(n-1)
两边同乘2
2Sn=2+3/1+5/2+……+(2n-1)/2^(n-2)
两式错位相减
2Sn-Sn=2+[2+2/2+2/4+……+2/2^(n-2)]-(2n-1)/2^(n-1)
Sn=2+2×(1-(1/2)^(n-1))/(1-1/2)-(2n-1)/2^(n-1)
=6-(2n+3)/2^(n-1)
2.
当n为偶数时n=2k,
A(2k)+A(2k-1)=(2(2k)+1)×(2(2k)-1)-(2(2k-1)+1)×(2(2k-1)-1)
=(4(2k)^2-1)-(4(2k-1)^2-1)
=4(4k-1)
Sn=S(2k)=[(A2+A1)+(A(2k)+A(2k-1))]×k/2
=(4×3+4(4k-1))×k/2
=4k(2k+1)
=2n(n+1)
当n为奇数时,
Sn=S(n-1)+An
S(n-1)用上面求得公式套
S(n-1)=2(n-1)((n-1)+1)=2n(n-1)
Sn=2n(n-1)+(-1)^n×(2n+1)×(2n-1)
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