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I was helping my brother with his maths homework, where he has just started learning about arithmetic series and their formulas such as the sum of the first $n$ terms ($S_n$) or finding the $n$th term of the sequence ($T_n$). While looking through some of the questions in his textbook, I came across this question:

"Find the original sequence $T_n$ if its partial sums $S_n$ are given by $S_n = n^2+5$"

We can find $T_n$ if we take $S_n-S_{n-1}$ :

$$S_n-S_{n-1}=n^2+5 - (n-1)^2-5=2n-1 $$

$$ \Rightarrow T_n =2n-1$$ which is the expected solution to the question by the textbook. However upon trying to find an actual sequence of numbers that fit the solution I came up empty handed! I feel as if there is no way the sum can have a constant term but I'm not sure why.

Is the textbook's solution actually correct? Is there such a series of numbers that works?

If the textbook's solution is wrong, am I correct in assuming that for any arithmetic series of numbers, the sum of the series cannot be of the form $S_n=f(n)+k$, where $k$ is a constant?

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  • $\begingroup$ I guess you can define the $0^{th}$ term as 5? $\endgroup$
    – stochs
    Commented Nov 18, 2023 at 7:22
  • $\begingroup$ For partial sums, the usual convention is $S_0 = \sum_{i = 1}^{0}T_i = 0$. So $S_0 = 5$ is impossible. $\endgroup$
    – Kakashi
    Commented Nov 18, 2023 at 21:21

1 Answer 1

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The textbook's answer is wrong.

$T_n=S_n-S_{n-1}$ is true only when $n\ge2$. It's not true when $n=1$, because then we would have $T_1=S_1-S_0$ but $S_0$ is not defined.

So for $n\ge2$ we have $T_n=2n-1$ as you showed. For $T_1$, we have $T_1=S_1=1^2+5=6$.

So the first few terms of the sequence are $6,3,5,7,9,\dots$. The sequence is not arithmetic. But if we ignore the first term, then it's arithmetic.

The sum of an arithmetic series is $S_n=\frac{n}{2}(T_1+(n-1)d)$ where $T_1$ is the first term, and $d$ is the common difference. This gives $S_n=\frac{d}{2}n^2+\frac12(T_1-d)n$. So if $S_n=f(n)+k$ (where $f(n)$ itself has no constant term) then $k$ must equal $0$.

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