一個有關級數的問題(太久沒碰)

1+3/2+5/4+7/8+9/16+.....+(2n－1)/2^(n－1)
Sol
A=1+3/2+5/4+7/8+9/16+.....+(2n－1)/2^(n－1)
A/2= 1/2+3/4+5/8+7/16+.....+(2n－3)/2^(n－1)+(2n－1)/2^n
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A/2=1+[2/2+2/4+2/8+2/16+…+2/2^(n－1)]－(2n－1)/2^n
=1+[2－2/2^(n－1)]－(2n－1)/2^n
=1+2－2/2^(n－1)－(2n－1)/2^n
=3－(4+2n－1)/2^n
=3－(3+2n)/2^n
A=6－(3+2n)/2^(n－1)

令Sn=1+(3/2)+(5/4)+(7/8)+(9/16)+.....+(2n-3/2^(n-2))+(2n-1/2^(n-

1))--------(1)


則2*Sn =2+3+(5/2)+(7/4)+(9/8)+-----(2n-1/2^(n-2))--------(2)


(2)-(1)得到

Sn=4+(2/2)+(2/4)+(2/8)+(2/16)+---- +(2/2^(n-2))-(2n-1/2^(n-1))

=5+ (1/2)+(1/4)+(1/8)+---+(1/2^(n-3)) -(2n-1/2^(n-1))

=5+ (1/2)[1-(1/2)^(n-3)]/[1-(1/2)] -(2n-1/2^(n-1))

=5+1-(1/2)^(n-3)-(2n-1/2^(n-1))

=6-(4/2^(n-1))-(2n-1/2^(n-1))

=6- [(2n+3)/2^(n-1)]