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

 Oct 28, 2019

Best Answer 

 #1
avatar+236 
+2

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.

 Oct 28, 2019
 #1
avatar+236 
+2
Best Answer

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.

ThatOnePerson Oct 28, 2019
 #2
avatar+23337 
+2

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

 

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

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

 

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

 

 

laugh

 Oct 29, 2019
edited by heureka  Oct 29, 2019
edited by heureka  Oct 29, 2019

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