If the difference sequence of {a_n} is defined by b_n = a_{n+1} − a_n, then
a_n = a_1 + Σ_{k=1}^{n−1} b_k (n ≧ 2)
Example ① Find the general term for a_1=1, a_{n+1}=a_n+2n.
Solution: Difference b_n = 2n.
a_n = 1 + Σ_{k=1}^{n−1} 2k = 1 + 2·(n−1)n/2 = n² − n + 1
Solution: Difference b_n = 2n.
a_n = 1 + Σ_{k=1}^{n−1} 2k = 1 + 2·(n−1)n/2 = n² − n + 1
Key Point
- Do not forget to check n=1: after deriving the result for n≧2, confirm that it agrees with a_1
Practice Problems
- Find the general term for a_1=2, a_{n+1}=a_n+3n+1.
Solutions and Explanations
- Solution:a_n = (3n²−n+2)/2
Add the difference 3n+1 using Σ, then add 2.