What is nth number on this series? I have series of
1+5+12+22+.....

where d=3n+1 where n=0,1,2...

but i can not find the nth number on this series ? can anyone explain how to get nth number on this series?
 A: $$1=1$$
$$5=1+(1+3(1))$$
$$12=1+(1+3(1))+(1+3(2))$$
etc.
If we collect all the ones, we get
$$a_k=\underbrace{1+1+\dots+1}_k+3(0+1+2+\dots+(k-1))$$
$$a_k=k+3T_{k-1}$$
where $T_k$ are the triangle numbers, giving the final solution of:

$$a_k=\frac{k(3k-1)}2$$

A: We have $$T_1 =1$$ $$T_2 =5 =1+4$$ $$T_3 = 12 = 1+4+(4+3) $$ $$T_4 =22 =1+4+ (4+3) + (4+3+3) $$ $$T_5 =35 = 1+4+(4+3)+(4+3+3)+(4+3+3+3) $$ $$\vdots $$
We can thus write $$T_n = 1 + 4 +(4+3) +(4+2\times 3) + \cdots +(4+(n-2)\times 3) $$ $$= 1 + (n-1)\times 4 + (1+2+3+\cdots + (n-2))\times 3$$ $$=1+(n-1)\times 4 + (\sum_{i=1}^{n-2} i)\times 3$$ $$=1+(4n-4) + 3(\frac {(n-2)(n-1)}{2})$$ $$ = (4n-3) + \frac {3n^2-9n+6}{2}$$ $$\boxed {T_n = \frac {3n^2-n}{2}} $$
Hope it helps. 
A: The successive differences are $$5-1,12-5,22-12$$
so, the difference between the $n$th term $T_n$ and the $T_{n-1}$th term of $1,5,12,22$ is $3n+1$
Let $T_m=a_0+a_1m+a_2m^2+\cdots$
So, $3n+1=T_n-T_{n-1}=a_1+a_2\{n^2-(n-1)^2\}+a_3\{n^3-(n-1)^3\}+\cdots$
Clearly, $a_r=0$ for $r\ge3$
Comparing the coefficients of $n,2a_2=3$
Comparing the constants $1=a_1-a_2\iff a_2=?$
$1=T_1=a_0+a_1+a_2\cdot1^2\iff a_0=?$
A: The successive differences are $$5-1,12-5,22-12$$
so, the difference between the $n$th term $T_n$ and the $T_{n-1}$th term of $1,5,12,22$ is $3n+1$
$$\implies T_n=T_1+\sum_{r=1}^{n-1}(3r+1) =1+\dfrac{(n-1)(4+3n-2)}2$$
A: Let a(n) denote the first in the sequence.
a(0) : 1
a(1): 5
Then a(n) - a(n-1) equals 3n+1. Or a(n) equals a(n-1) + 3n + 1. 
Note that sigma( i=0, inf): n = n(n+1)/2. 
Also, note there are n times the second term in d. 
Then the closed form a(n( with n >=1  = 3(n(n+1)/2)+1+n.
Sorry if I can't use standard notation tools. I'm blind and those tools can't be used on my screen reader
