1. ## series

the natural numbers are grouped as follows:
(1) , ( 2 , 3 ) , ( 4 , 5 , 6 ) , ( 7 , 8 , 9 , 10 ) , ...

find an expression for the first term of the nth group and prove that the sum of the numbers in the nth group is 1/2 n (n^2+1).

uh...I have no idea where to even start.

2. Originally Posted by colloquial
the natural numbers are grouped as follows:
(1) , ( 2 , 3 ) , ( 4 , 5 , 6 ) , ( 7 , 8 , 9 , 10 ) , ...

find an expression for the first term of the nth group and prove that the sum of the numbers in the nth group is 1/2 n (n^2+1).

uh...I have no idea where to even start.
Here is a pattern i found. I hope it helps.

3. The nth grouping will be $\left( {\frac{{n^2 - n + 2}}{2}, \cdots ,\frac{{n^2 + n}}{2}} \right)$.
Now apply the Gauss formula for the sum of the first n positive integers.

4. Hello, colloquial!

The natural numbers are grouped as follows:
$\{1\},\;\{2,3\},\;\{4,5,6\},\;\{7,8,9,10\},\;...$

(a) Find an expression for the first term of the $n^{th}$group.

(b) Prove that the sum of the numbers in the $n^{th}$ group is: . $\frac{1}{2}n(n^2+1)$

Note the last number of each group: . $1,\,3,\,6,\,10,\,\cdots$
. . These are the triangular numbers: . $T_n \:=\:\frac{n(n+1)}{2}$
So each group begins with the preceding triangular number, plus 1.

(a) The first term of the $n^{th}$ group is: . $a_n \:=\:\frac{n(n-1)}{2}+1\;=\;{\color{blue}\frac{n^2-n+2}{2}}$

(b) Each group is an arithmetic series with first term, $a_1 \:=\:\frac{n^2-n+2}{2}$,
. . common difference, $d = 1$, and $n$ terms.

The sum of an arithmetic series is: . $S_n \;=\;\frac{n}{2}\left[2a_1 + (n-1)d\right]$

Therefore: . $S_n \;=\;\frac{n}{2}\left[2\!\cdot\!\frac{n^2-n+2}{2} + (n-1)\!\cdot\!1\right] \;=\;{\color{blue}\frac{1}{2}n(n^2+1)}$