Question

4.已知数列{an}满足 a(n+1)=3an, 且 a1=-1, 则数列 {an+2n} 的前5？

Answer 1

首先，我们可以利用递推式 $a_{n+1} = 3a_n$ 来计算数列的前几项：
a
2
=
3
a
1
=
3
(
−
1
)
=
−
3
a
3
=
3
a
2
=
3
(
−
3
)
=
−
9
a
4
=
3
a
3
=
3
(
−
9
)
=
−
27
a
5
=
3
a
4
=
3
(
−
27
)
=
−
81
a
6
=
3
a
5
=
3
(
−
81
)
=
−
243
a
7
=
3
a
6
=
3
(
−
243
)
=
−
729

a
2
​

a
3
​

a
4
​

a
5
​

a
6
​

a
7
​

​

=3a
1
​
=3(−1)=−3
=3a
2
​
=3(−3)=−9
=3a
3
​
=3(−9)=−27
=3a
4
​
=3(−27)=−81
=3a
5
​
=3(−81)=−243
=3a
6
​
=3(−243)=−729
​

现在我们考虑数列 ${a_{n+2n}}$，即将数列 ${a_n}$ 中每个数加上其下标的两倍，得到的新数列。
a
3
+
2
×
1
=
−
9
+
2
=
−
7
a
4
+
2
×
2
=
−
27
+
4
=
−
23
a
5
+
2
×
3
=
−
81
+
6
=
−
75
a
6
+
2
×
4
=
−
243
+
8
=
−
235
a
7
+
2
×
5
=
−
729
+
10
=
−
719

a
3
​
+2×1
a
4
​
+2×2
a
5
​
+2×3
a
6
​
+2×4
a
7
​
+2×5
​

=−9+2=−7
=−27+4=−23
=−81+6=−75
=−243+8=−235
=−729+10=−719
​

因此，数列 ${a_{n+2n}}$ 的前 $5$ 项为 $-7, -23, -75, -235, -719$。