How to express the following as a series? How you you express $$1+3+6+10+15+21+28+...$$ as a summation? I've tried different ways but they were wrong and always brought me back to square one. Thanks!
 A: $$\sum_{i=1}^\infty \frac{i(i+1)}{2}$$
Why: we want the $i^\textrm{th}$ term to be $1+2+\cdots+i$.
A: For the $nth$ term $S(n)$ of your sum, we have $\displaystyle S(n) = \sum_{k = 1}^n k = \dfrac{n(n+1)}{2}$. 
If you want the sum of the first $n$ such terms, $S(1) + S(2) + ... + S(n)$ use:
$$\sum_{k = 1}^n \dfrac{k(k+1)}{2} = \dfrac{n(n+1)(n+2)}{6}$$
See the link to triangular numbers to learn more!
A: You can write the sequence being summed as a recurrence relation for $n \geq 2$, assuming we index the sequence as $(a_1,a_2,a_3,\cdots)$, as:
$$a_n=a_{n-1}+n.$$
Recurrence relations of this form with $a_n=a_{n-1}+p(n)$ for $p(n)$ a polynomial, where the degree of the polynomial is say $m \geq 0$, have $a_n$ of the form:
$$a_n=c_0+c_1n+\cdots +c_mn^m+c_{m+1}n^{m+1}.$$
Therefore, our general term $a_n$ will look like
$$a_n=c_0+c_1n+c_2n^2$$
and we can use the information that $a_1=1,a_2=3$ and $a_3=6$ to get
$$a_1=c_0+c_1+c_2=1$$
$$a_2=c_0+2c_1+4c_2=3$$
$$a_3=c_0+3c_2+9c_2=6.$$
We then solve the above system of linear equations to get that $c_0=0,c_1=\frac{1}{2},$ and $c_2=\frac{1}{2}$ which gives upon simplification that
$$a_n=\frac{n(n+1)}{2}=T_n$$
the $n^{th}$ triangular number. Thus finally,
$$1+3+6+10+15+\cdots = \sum_{n=1}^{\infty} \frac{n(n+1)}{2}.$$
A: $$S_n=1+3+6+10+15+\cdots+T_n$$
$$S_n=1+3+6+10+15+\cdots+T_{n-1}+T_n$$
On subtraction, $$0=1+(3-1)+(6-3)+(10-6)+(15-10)+\cdots+(T_n-T_{n-1})-T_n$$
$$\implies T_n=1+2+3+\cdots\text{up to }n\text{ terms}=\frac{n(n+1)}2$$
$$\text{Now, }(r+1)^3-r^3= 6\frac{r(r+1)}2+1$$ (Observe the Telescoping Sum in the Left Hand Side)
Putting $r=1,2,\cdots,n-1,n$ and adding them we get, 
 $$(n+1)^3-1=6S_n+n$$
$$\implies 6S_n=n^3+3n^2+3n+1-1-n=n^3+3n^2+2n=n(n+1)(n+2)$$
