52577412
An arithmetic sequence has first term \(a_1=5\) and common difference \(d=2\).
a) Find a closed formula for the partial sum \(S_n=\sum_{k=1}^{n}a_k\).
b) Determine whether \(b_n=\frac{S_n}{n^2}\) converges, and find its limit as \(n\to\infty\).
c) Show algebraically that
\(1+2+\cdots+(n-1)+n+(n-1)+\cdots+2+1=n^2\)
for every positive integer \(n\).
Hints
- Find the explicit formula for the arithmetic sequence first.
- Simplify \(b_n\) so its limiting behavior is visible.
- Split the symmetric sum into two sums of consecutive positive integers.
Solution
1. The explicit formula for the arithmetic sequence is
\(a_n=5+2(n-1)=2n+3\).
Therefore,
\(S_n=\frac n2(a_1+a_n)=\frac n2[5+(2n+3)]=n^2+4n\).
2. Then
\(b_n=\frac{n^2+4n}{n^2}=1+\frac4n\).
Since \(\frac4n\to0\), the sequence converges to \(1\).
3. Split the symmetric sum into two familiar sums:
\(\frac{n(n+1)}{2}+\frac{(n-1)n}{2}=\frac{n^2+n+n^2-n}{2}=n^2\).
Answer
a) \(S_n=n^2+4n\).
b) The sequence converges to \(1\).
c) \(\frac{n(n+1)}{2}+\frac{(n-1)n}{2}=n^2\).
