Calculating the $100th$ term of a series I came across a series
$1,5,13,25,41,61.......$
I have to calculate the $100^{th}$ term of this series. How do I do it?
Is there any formula to calculate the $n^{th}$ term of a series?
 A: Looks to me as though the forward difference is linear so that your equation can be given by a quadratic.
$$4,8,12,..$$
It seems as though,
$$f(x+1)-f(x)=4x+4$$
Where $f(0)=1$ (and $f(1)=5$).
So by summing both sides of this equation from $x=0$ to $n-1$ and noticing that the sum in the LHS telescopes we have,
$$f(n)-f(0)=\sum_{x=0}^{n-1} (4x+4)$$
$$f(n)=f(0)+\sum_{x=0}^{n-1} (4x+4)$$
$$f(n)=1+4\frac{(n-1)(n)}{2}+4n$$
Now if instead you want the sequence to start with $f(1)$ instead of $f(0)$, shift current $f$ to the right $1$.
$$f(n)=1+2(n-1)((n-1)-1)+4(n-1)$$
$$f(n)=2n^2-2n+1$$
So,
$$f(100)=2(100^2)-2(100)+1$$
$$=19801$$
A: The difference of successive terms forms an Arithmetic progression: $4,8,\cdots $. 
We can write the terms of our sequence as: $$0\times 4+1,1\times 4+1, 3\times 4+1, 6\times 4+1, 10\times 4+1, \cdots $$ $$=\frac {0\times 1}{2}\times 4+1, \frac {1\times 2}{2}\times 4+1, \frac {2\times 3}{2}\times 4+1, \frac {3\times 4}{2}\times 4+1, \cdots $$
We can see a pattern emerging. Thus the $n$th term of the sequence is $$\frac {n (n-1)}{2}\times 4+1 =4\binom {n}{2} +1$$ The hundredth term is thus $\boxed {4\binom {100}{2}+1=19801}$. Hope it helps. 
A: In general, there is no way to determine the $n$th term of a series.
For this series 1, 5, 13, 25, 41, 61, ...
$1 + 4\times 1 = 5$ 
$5 + 4\times 2 = 13$
$13 + 4\times 3 = 25$
...
$\Rightarrow f(n)=f(n-1)+4(n-1), f(0)=1 \Rightarrow f(n)=(4\sum_{i=1}^{n-1}i)+1 $
$f(100)=(4\sum_{i=1}^{99}i)+1=19801$
A: Your sequence is $1+2n(n+1)$, so the $100^{th}$ term is $20201$.
