help

Assuming that you want us to answer in terms of n, the first few numbers of this sequence are 1,3,6,10... those are the triangular numbers. We know that the formula for finding the sum of the first n triangular numbers is n(n+1)/2, so that's your answer.

Find 1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ... + (1 + 2 + 3 + ... + n).

\(\mathbf{s_n=\ ?}\)

\(\begin{array}{|lcccccccccc|} \hline \text{First row}: & \color{red}d_0=1 & &3 & &6 & &10 & & 15 & \ldots \\ \text{Second Row}: & &\color{red}d_1=2 & &3 & &4 & &5 & \ldots \\ \text{Third row}: & & &\color{red}d_2=1 & &1 & & 1& \ldots \\ \hline \end{array} \)

\(\begin{array}{rcl} s_n &=& \dbinom{n}{1}\cdot {\color{red}d_0 } + \dbinom{n}{2}\cdot {\color{red}d_1 } + \dbinom{n}{3}\cdot {\color{red}d_2 }\\ s_n &=& \dbinom{n}{1}\cdot {\color{red}1 } + \dbinom{n}{2}\cdot {\color{red}2} + \dbinom{n}{3}\cdot {\color{red}1}\\ \\ \hline \binom{n}{1} &=& n \\ \binom{n}{2} &=& ( \dfrac{n}{2} ) \cdot ( \dfrac{n-1}{1} ) \\ \binom{n}{3} &=& ( \dfrac{n}{3} ) \cdot ( \dfrac{n-1}{2} )\cdot ( \frac{n-2}{1} ) \\ \hline \\ s_n &=& (n)\cdot {\color{red}1} + ( \dfrac{n}{2} ) \cdot ( \dfrac{n-1}{1} )\cdot {\color{red}2} + ( \dfrac{n}{3} ) \cdot ( \dfrac{n-1}{2} )\cdot ( \dfrac{n-2}{1} )\cdot {\color{red}1} \quad | \quad \cdot 6\\ \\ 6\cdot s_n &=& n\cdot 6 + n \cdot ( n-1 )\cdot 6 + n \cdot ( n-1 )\cdot ( n-2 ) \\ 6\cdot s_n &=& n \left[~ 6 + ( n-1 )\cdot 6 + ( n-1 )\cdot ( n-2 ) ~\right] \\ 6\cdot s_n &=& (n) \left(~ 6 + 6n-6 + n^2 - 3n + 2 ~\right) \\ 6\cdot s_n &=& (n) \left(~ n^2 + 3n + 2 ~\right) \\ 6\cdot s_n &=& (n) \cdot (n+1) \cdot ( n+2 ) \\\\ \mathbf{s_n} &=& \mathbf{ \dfrac{ n \cdot (n+1) \cdot ( n+2 ) }{6} } \qquad \text{or} \qquad \mathbf{s_n} = \dbinom{n+2}{3} \\\\ s_1 &=& 1 = \dfrac{ 1 \cdot 2 \cdot 3}{6} = 1\\ s_2 &=& 1+3 = \dfrac{ 2 \cdot 3 \cdot 4 }{6} = 4 \\ s_3 &=& 1+3+6 = \dfrac{ 3 \cdot 4 \cdot 5 }{6} = 10 \\ s_4 &=& 1+3+6+10 = \dfrac{ 4 \cdot 5 \cdot 6 }{6} = 20\\ s_5 &=& 1+3+6+10+15 = \dfrac{ 5 \cdot 6 \cdot 7 }{6} = 35\\ \cdots \end{array}\)