问题
选择题
数列{an},a1=1,an+an+1=2n,则数列{an+1-an}的前10项和T10=( )
A.0
B.5
C.10
D.20
答案
∵a1=1,an+an+1=2n,
∴a2=1,a3=3,a4=3,a5=5,a6=5,a7=7,a8=7,a9=9,a10=9,
∴数列{an+1-an}的前10项和T10=a2-a1+a3-a2+…a10-a9=a9+a2=10,
故选C.
数列{an},a1=1,an+an+1=2n,则数列{an+1-an}的前10项和T10=( )
A.0
B.5
C.10
D.20
∵a1=1,an+an+1=2n,
∴a2=1,a3=3,a4=3,a5=5,a6=5,a7=7,a8=7,a9=9,a10=9,
∴数列{an+1-an}的前10项和T10=a2-a1+a3-a2+…a10-a9=a9+a2=10,
故选C.