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(1)已知等差数列{an},bn=a1+a2+a3+…+ann(n∈N*),求证:{bn}仍为等差数列;(2)已知等比数列{cn},cn>0(n∈N*)),类比上述性质,写出一

题目详情
(1)已知等差数列{a n }, b n =
a 1 + a 2 + a 3 +…+ a n
n
(n∈N * ),求证:{b n }仍为等差数列;
(2)已知等比数列{c n },c n >0(n∈N * )),类比上述性质,写出一个真命题并加以证明.
▼优质解答
答案和解析
(1)由题意可知b n =
n( a 1 + a n )
2
n
=
a 1 + a n
2

∴b n+1 -b n =
a 1 + a n+1
2
-
a 1 + a n
2
=
a n+1 - a n
2

∵{a n }等差数列,∴b n+1 -b n =
a n+1 - a n
2
=
d
2
为常数,(d为公差)
∴{b n }仍为等差数列;
(2)类比命题:若{c n }为等比数列,c n >0,(n∈N * ),
d n =
n c 1 • c 2 … c n
,则{d n }为等比数列,
证明:由等比数列的性质可得:d n =
n ( c 1 c n )
n
2
=
c 1 c n

d n+1
d n
=
c n+1
c n
=
q
为常数,(q为公比)
故{d n }为等比数列