For example:
an=2n- 1、bn=( 1/2)^(n)
Let: cn = anbn = (2n-1) × (1/2) n.
Then the sum of the first n terms of the sequence {cn} is t n, so:
TN = 1×( 1/2)+3×( 1/2)? +5×( 1/2)? +…+(2n- 1)×( 1/2)^n
( 1/2)Tn = = = = = = = 1×( 1/2)? +3×( 1/2)? +…+(2n-3)×( 1/2)^n+(2n- 1)×( 1/2)^(n+ 1)
When subtracting these two types, please pay attention to the in braces, so:
( 1/2)Tn = 1×( 1/2)+2×( 1/2)? +2×( 1/2)? +…+2×( 1/2)^n-(2n- 1)×( 1/2)^(n+ 1)
What is in braces can be summarized by geometric series.
( 1/2)tn=( 1/2)+ 1-(2n+3)×( 1/2)^(n+ 1)
Get:
Tn=3-(2n+3)×( 1/2)^n