Why is my method of finding $A_7$ wrong?

Question

If $S_n$ represents the sum of first $n$ terms of an A.P. defined as, $$S_n=\frac{n^2+1}{n+1}$$

I need to calculate $7$th term of the A.P. i.e. $A_7$.

So I found $S_{n-1}$ and subtracted it from $S_n$ and then plugged in $7$, I was incorrect.

My formula was $\frac{n-n^2}{n^2+n}$ from $\frac{n^2+1}{n+1}-\frac{n^2}{n}$

What exactly did I end up calculating? I know the way which I'm supposed to answer the question, I thought that what I did should have given the same answer though.

Also what symbol should I use for subscript here? Thankyou.

Answer 1

That's just bad algebra! With $S_n= \frac{n^2+ 1}{n+ 1}$, $S_{n-1}= \frac{(n-1)^2+ 1}{(n-1)+ 1}= \frac{n^2- 2n+ 2}{n}$ NOT "$\frac{n^2}{n}$.

Answer 2

But how about this? It cannot be that the first $n$ terms of an arithmetic progression (AP) sum to $(n^{2}+1)/(n+1)$.

Let $t_{n}$ be the $n^{th}$ term of an AP then we can write it as $t_{n}=a+(n-1)b$ where $b$ is the common difference.

So the sum of the first n terms will be $an-bn+n(n+1)b/2$. There are no values of $a$ and $b$ that can make this into $(n^{2}+1)/(n+1)$.

Or let's look at it another way. The first $10$ values of $S_{n}=(n^{2}+1)/(n+1)$ are:

$1,5/3,5/2,17/5,13/3,37/7,25/4,65/9,41/5,101/11$ If these are the sums then the terms are:

$1,2/3,5/6,9/10,14/15,....$

If this was an arithmetic progression then it would have a common difference. The first two terms suggest it would be $1/3$ and terms two and three suggest that it would be $1/6$. So this is not an AP and we should just refer to $S_{n}$ as the sum of a sequence.