Question

等差數(shù)列有如下性質(zhì)，若數(shù)列{a_n}是等差數(shù)列，則當(dāng)bn=

a1+a2+…+an

n

(n∈N*)時(shí)，數(shù)列{b_n}也是等差數(shù)列；類比上述性質(zhì)，相應(yīng)地{c_n}是正項(xiàng)等比數(shù)列，當(dāng)數(shù)列d_n=

(c1c2…cn)

1

n

時(shí)，數(shù)列{d_n}也是等比數(shù)列．

Answer 1

分析：數(shù)列{a_n}是等差數(shù)列，則當(dāng)bn=

a1+a2+…+an

n

(n∈N*)時(shí)，數(shù)列{b_n}也是等差數(shù)列說明等差數(shù)列的前n項(xiàng)和除以項(xiàng)數(shù)構(gòu)成新的等差數(shù)列，由此類比，數(shù)列{c_n}是正項(xiàng)等比數(shù)列，則它的前n項(xiàng)的乘積開項(xiàng)數(shù)次方也構(gòu)成新的等比數(shù)列，然后利用等比數(shù)列的定義加以證明．

解答：解：由數(shù)列{a_n}是等差數(shù)列，則當(dāng)bn=

a1+a2+…+an

n

(n∈N*)時(shí)，數(shù)列{b_n}也是等差數(shù)列．
類比得到：{c_n}是正項(xiàng)等比數(shù)列，當(dāng)數(shù)列dn=(c1c2…cn)

1

n

時(shí)，數(shù)列{d_n}也是等比數(shù)列．
證明如下：
∵{c_n}是正項(xiàng)等比數(shù)列，設(shè)其公比為q，
∴(c1c2…cn)

1

n

=(c1nq1+2+…+n-1)

1

n

=c1q

n-1

2

．
(c1c2…cn-1)

1

n-1

=(c1n-1q1+2+…+n-2)

1

n-1

=c1q

n-2

2

．
∴

dn

dn-1

=

(c1c2…cn) 1 n

(c1c2…cn-1) 1 n-1

=

c1q n-1 2

c1q n-2 2

=q

1

2

．
∴當(dāng)數(shù)列dn=(c1c2…cn)

1

n

時(shí)，數(shù)列{d_n}也是等比數(shù)列．
故答案為：(c1c2…cn)

1

n

．

點(diǎn)評(píng)：本題考查了等比關(guān)系的確定，考查了類比推理，是中檔題．