数学中错项相减求解Sn为数列{an}前n项和,a1=10,a2=8,an=12-2n.1)若bn=an乘以2的n次方,求{|bn|}的前n项和Tn.怎样用错位相消来做望详解an为等差数列

Question

数学中错项相减求解
Sn为数列{an}前n项和,a1=10,a2=8,an=12-2n.1)若bn=an乘以2的n次方,求{|bn|}的前n项和Tn.怎样用错位相消来做望详解
an为等差数列

Answer 1

T（n)=10x2+8x4+6x8+.+(14-2n)x2^(n-1)+(12-2n)x2^n
2T（n)= 10x4+8x8+.+(14-2n)x2^n+(12-2n)x2^(n+1)
二式-一式得：
T（n)=-10x2+2x4+2x8+.+2x2^n+(12-2n)x2^(n+1)
=-20+2（4+8+.+2^n)+(12-2n)x2^(n+1)
=-20+2x[2^(n+1)-4]+(12-2n)x2^(n+1)
=-28+4x2^n+(12-2n)x2^(n+1).
=(28-4n)x2^n -28.