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- 已知数列 {an}\left\{a_{n}\right\} 满足 a1=2a_{1}=2, 且 an+1=an2+4an+2a_{n+1}=a_{n}^{2}+4 a_{n}+2, 则 $...
Mar 8, 2024
- 已知数列 {an}\left\{a_{n}\right\} 满足 a1=2a_{1}=2, 且 an+1=an2+4an+2a_{n+1}=a_{n}^{2}+4 a_{n}+2, 则 an=a_{n}= ; 令 bn=1an+3+1an−1+1b_{n}=\frac{1}{a_{n}+3}+\frac{1}{a_{n-1}+1} ，若 {bn}\left\{b_{n}\right\} 的前 nn 项和为 SnS_{n}, 则 Sn=S_{n}=
Solution by Steps
step 1
To find the general term ana_n of the sequence, we observe the recursive formula an+1=an2+4an+2a_{n+1}=a_{n}^{2}+4a_{n}+2
step 2
Notice that the recursive formula can be rewritten as an+1=(an+2)2−2a_{n+1}=(a_n+2)^2-2, which suggests a substitution might simplify the sequence
step 3
Let's define a new sequence cn=an+2c_n = a_n + 2. Then cn+1=an+1+2=(an+2)2=cn2c_{n+1} = a_{n+1} + 2 = (a_n+2)^2 = c_n^2
step 4
Since a1=2a_1 = 2, we have c1=a1+2=4c_1 = a_1 + 2 = 4. The new sequence cnc_n is a geometric sequence with c1=4c_1 = 4 and common ratio cn/cn−1=cn−1c_n / c_{n-1} = c_{n-1}
step 5
The general term for cnc_n is cn=42n−1c_n = 4^{2^{n-1}}
step 6
Therefore, the general term for ana_n is an=cn−2=42n−1−2a_n = c_n - 2 = 4^{2^{n-1}} - 2
step 7
To find bnb_n, we use the definition bn=1an+3+1an−1+1b_n = \frac{1}{a_n+3} + \frac{1}{a_{n-1}+1}
step 8
Substitute ana_n and an−1a_{n-1} with their expressions: bn=142n−1+1+142n−2−1b_n = \frac{1}{4^{2^{n-1}}+1} + \frac{1}{4^{2^{n-2}}-1}
step 9
Notice that bnb_n simplifies to bn=42n−2−142n−1−1+42n−1+142n−1+1=1b_n = \frac{4^{2^{n-2}}-1}{4^{2^{n-1}}-1} + \frac{4^{2^{n-1}}+1}{4^{2^{n-1}}+1} = 1
step 10
Since each term bn=1b_n = 1, the sum of the first nn terms SnS_n is simply nn
Answer
an=42n−1−2a_n = 4^{2^{n-1}} - 2, Sn=nS_n = n
Key Concept
Geometric sequence and series summation
Explanation
The key to solving the problem is recognizing the recursive formula's pattern and simplifying it by a substitution that reveals a geometric sequence. The sum of the first nn terms of the sequence {bn}\{b_n\} is straightforward once we find that each term bnb_n equals 1.
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