How is $S_{n-1} = n^2 +2$?

Question

I'm learning Arithmetic Progression. There's an example given in my book which I'm not able to understand from yesterday. The example is:

If the sum to $n$ terms of a sequence is given by $S_n =n^2+2n+3$, find $t_n$ and hence find $t_1$ and $t_2$.

Solution:

$S_n =(n^2+2n+3)$ and $S_{n-1} = (n-1)^2 + 2(n-1)+3 = n^2+2$.

Therefore, $t_n = (S_n - S_{n-1}) = (n^2 +2n + 3) - (n^2 + 2) = (2n+1)$.

Consequently, $t_1 = 3$ and $t_2 =5$.

I'm not getting the first line of the solution. To be honest, I'm not able to understand the question. Please help. Sorry if I asked something foolish (as I'm quite week in mathematics) and also for my bad English.

Thanks in advance!

Answer 1

First just replace $n$ by $n-1$ then expand $S_{n-1} = (n-1)^2 + 2(n-1) + 3$ to get

$$\require{cancel} S_{n-1} = n^2 - \cancel{2n} + 1 + \cancel{2n} - 2 + 3 = n^2 + 2$$

Then $t_n = S_n -S_{n-1}$ is a consequence.

Edit:

If you consider a sequence $(t_n)_{n \in \mathbb N}$ then a sum of the first $n$ terms is represented by

$$S_n = t_1 + t_2 + \ldots + t_n$$

As long as it was given the sum $S_n = n^2 + 2n + 3$, you are asked to find the terms of your sequence $(t_n)_{n \in \mathbb N}$, which follows from the algebra above.

Answer 2

The question is asking you this:

We have some sequence $t_n$ which looks like this:

$$t_1,t_2,t_3,\dots$$

And we won't tell you what $t_n$ is. But, we can tell you that we call the sum of the first $n$ terms of the sequence $S_n$, so

$$S_1=t_1\\ S_2=t_1+t_2\\ S_3=t_1+t_2+t_3\\ \vdots$$

And we can also tell you that, for every $n$, the value of $S_n$ is

$$S_n=n^2+2n+3$$

Now, can you figure out what $t_n$ is?

The answer is yes, because $$S_n - S_{n-1} = (t_1+t_2+\cdots+t_{n-1}+ t_n) - (t_1+t_2+\cdots+t_{n-1})$$

meaning that $S_n- S_{n-1} = t_n$. Now, since $S_n = n^2 + 2n + 3$ and $S_{n-1} = (n-1)^2 + 2(n-1) + 3$, calculating $t_n$ is easy.

Answer 3

$$S_n=t_1+t_2+t_3+...+t_{n-2}+t_{n-1}+t_n$$ $$S_{n-1}=t_1+t_2+t_3+...+t_{n-2}+t_{n-1}$$

Subtracting,we get $S_n-S_{n-1}=t_n$

Thus,$t_n=(2n+1)$

Now,just put n=1 and n=2

Answer 4

What is missed out by the book solution is that the sum does not work.

In order to fix that, we need the first term - whether it is (as implied) $t_1$ or potentially $t_0$ - to be different from the formula which we can use for the other terms. For the first term, there is no $S_{n-1}$, and we must have that $t_n$=$S_n$ (for the first term only).

So, assuming that $t_1$ is the first term and there is no $t_0$, we get that $t_1 = 1^2 + 2\times 1 +3 = 6$.

Or if we have an extra $t_0$ that is the first term, then $t_0=S_0=3$