解题：原式＝
2. 在递增的等比数列
(
a
n
)
(a_n)
(an)中，若
(
a
3
−
a
1
=
5
2
)
(a_3 - a_1 = \frac{5}{2})
(a3−a1=25),
(
a
2
=
3
)
(a_2 = 3)
(a2=3), 则公比 (q) =
A.
(
4
3
)
( \frac{4}{3} )
(34)
B.
(
3
2
)
( \frac{3}{2} )
(23)
C. 2
D.
(
5
2
)
( \frac{5}{2} )
(25)
为找到公比q，
[
a
n
=
a
1
⋅
q
(
n
−
1
)
]
[ a_n = a_1 \cdot q^{(n-1)} ]
[an=a1⋅q(n−1)]
[
a
3
=
a
1
⋅
q
(
3
−
1
)
=
a
1
⋅
q
2
]
[ a_3 = a_1 \cdot q^{(3-1)} = a_1 \cdot q^2 ]
[a3=a1⋅q(3−1)=a1⋅q2];
[
a
2
=
a
1
⋅
q
(
2
−
1
)
=
a
1
⋅
q
]
[ a_2 = a_1 \cdot q^{(2-1)} = a_1 \cdot q ]
[a2=a1⋅q(2−1)=a1⋅q]
所以有：
[
a
3
−
a
1
=
5
2
]
[
a
1
⋅
q
2
−
a
1
=
5
2
]
[
a
1
⋅
(
q
2
−
1
)
=
5
2
]
[
q
2
−
1
=
5
2
a
1
]
[
q
2
=
5
2
a
1
+
1
]
[
q
2
=
5
+
2
a
1
2
a
1
]
[ a_3 - a_1 = \frac{5}{2} ] [ a_1 \cdot q^2 - a_1 = \frac{5}{2} ] [ a_1 \cdot (q^2 - 1) = \frac{5}{2} ] [ q^2 - 1 = \frac{5}{2a_1} ] [ q^2 = \frac{5}{2a_1} + 1 ] [ q^2 = \frac{5 + 2a_1}{2a_1} ]
[a3−a1=25][a1⋅q2−a1=25][a1⋅(q2−1)=25][q2−1=2a15][q2=2a15+1][q2=2a15+2a1]
[ q = a 2 a 1 = 3 a 1 ] [ q = \frac{a_2}{a_1} = \frac{3}{a_1} ] [q=a1a2=a13]
[ ( 3 a 1 ) 2 = 5 + 2 a 1 2 a 1 ] [ \left(\frac{3}{a_1}\right)^2 = \frac{5 + 2a_1}{2a_1} ] [(a13)2=2a15+2a1]
[ 9 = 5 + 2 a 1 2 a 1 ] [ 18 a 1 = 5 + 2 a 1 ] [ 16 a 1 = 5 ] [ a 1 = 5 16 ] [ 9 = \frac{5 + 2a_1}{2a_1} ] [ 18a_1 = 5 + 2a_1 ] [ 16a_1 = 5 ] [ a_1 = \frac{5}{16} ] [9=2a15+2a1][18a1=5+2a1][16a1=5][a1=165]
[ q = 3 a 1 = 3 5 16 = 48 5 = 24 5 = 4 1 = 4 ] [ q = \frac{3}{a_1} = \frac{3}{\frac{5}{16}} = \frac{48}{5} = \frac{24}{5} = \frac{4}{1} = 4 ] [q=a13=1653=548=524=14=4]
