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

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.

• You should really use MathJax as while HTML tags will work, MathJax will format your math much better (and it's less of a pain than typing out <sup> and <sub> over and over again as opposed to ^ and _). Feb 10, 2015 at 12:26
• Thanks for your kind advice! :) Feb 10, 2015 at 12:38

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.

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.

• how S<sub>n-1</sub> is equal to (n -1)<sup>2</sup> + 2(n-1) + 3 ? Feb 10, 2015 at 12:24
• It's a partial sum, which means that $S_{n-1} = t_1 + t_2 + \ldots + t_{n-1}$. Feb 10, 2015 at 12:27
• Can u please give me some advice so that i can strong my arithmetic. Feb 10, 2015 at 12:37
• I believe you are on the right track coming here to this site, a tip that I could give is practice, a whole lot! There is also this website that you could check, there is some nice stuff there. Also get yourself a good albegra book, it's very important. Feb 10, 2015 at 12:42

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

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

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

Now,just put n=1 and n=2

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 $t_0=S_0=3$