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\(\begin{array}{ccccccccccc}  \\ & & & & & 1   \\ & & & & 2 & & 2   \\ & & & 3 & & 4 & & 3   \\ & & 4 & & 7 & & 7 & & 4  \\ & 5 & & 11 & & 14 & & 11 & & 5  \\ 6 & & 16 & & 25 & & 25 & & 16 & & 6 \end{array} \)

If we form a number triangle as above, what is the sum of all the numbers in the triangle with 20 rows? 

 Nov 2, 2019

Best Answer 

 #1
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+1

This is the closed form of your sequence:


a_n = 1/2 (3 2^n - 4)......and it generates all the terms of your sequence as follows:


Sum = (1, 4, 10, 22, 46, 94, 190, 382, 766, 1534, 3070, 6142, 12286, 24574, 49150, 98302, 196606, 393214, 786430, 1572862) = 3,145,685 - which is the sum of the first 20 rows.

 Nov 2, 2019
 #1
avatar
+1
Best Answer

This is the closed form of your sequence:


a_n = 1/2 (3 2^n - 4)......and it generates all the terms of your sequence as follows:


Sum = (1, 4, 10, 22, 46, 94, 190, 382, 766, 1534, 3070, 6142, 12286, 24574, 49150, 98302, 196606, 393214, 786430, 1572862) = 3,145,685 - which is the sum of the first 20 rows.

Guest Nov 2, 2019
 #2
avatar+23893 
+1

help

\(\begin{array}{lccccccccccc} \text{row }1:& & & & & & 1 \\ \text{row }2:& & & & & 2 & & 2 \\ \text{row }3:& & & & 3 & & 4 & & 3 \\ \text{row }4:& & & 4 & & 7 & & 7 & & 4 \\ \text{row }5:& & 5 & & 11 & & 14 & & 11 & & 5 \\ \text{row }6:& 6 & & 16 & & 25 & & 25 & & 16 & & 6 \end{array} \)

If we form a number triangle as above,

what is the sum of all the numbers in the triangle with 20 rows?

 

\(\text{Let row $=r$} \\ \text{Let the sum of all the numbers in the rth row $=s_r$ } \\ \text{Let the sum of the numbers in the triangle from 1 to rth row $=s_{1\ldots r}$ }\)

 

\(\begin{array}{|rcll|} \hline s_1 &=& 1 \\ s_2 &=& 1 +3\cdot( 1 ) \\ s_3 &=& 1 +3\cdot( 1+2 ) \\ s_4 &=& 1 +3\cdot( 1+2+4 ) \\ s_5 &=& 1 +3\cdot( 1+2+4+8 ) \\ s_6 &=& 1 +3\cdot( 1+2+4+8+16 ) \\ \ldots \\ s_r &=& 1 +3\cdot( 1+2^1+2^2+2^3+2^4+\ldots + 2^{r-2} ) \\ &=& 1+ 3\cdot(2^{r-1}-1) \\ \mathbf{s_r} &=& \mathbf{ 3\cdot 2^{r-1} -2 } \\ \hline \end{array} \)

 

 

\(\begin{array}{|rcll|} \hline s_{1\ldots r} &=& \sum \limits_{i=1}^{r} \left(3\cdot 2^{i-1} -2 \right) \\ &=& 3\sum \limits_{i=1}^{r} \left(2^{i-1}\right) -\sum \limits_{i=1}^{r} 2 \\ &=& 3\sum \limits_{i=0}^{r-1} \left(2^i\right) -2\sum \limits_{i=1}^{r} 1 \\ &=& 3\sum \limits_{i=0}^{r-1} \left(2^i\right) -2r \\ &=& 3\sum \limits_{i=0}^{r} \left(2^i\right) -3\cdot2^r -2r \quad | \quad \sum \limits_{i=0}^{r} \left(2^i\right) = \left( 2^{r+1}-1 \right) \\ &=& 3\cdot \left( 2^{r+1}-1 \right) -3\cdot2^r -2r \\ &=& 3\cdot 2^{r+1}-1 -3 -3\cdot2^r -2r \\ &=& 3\cdot \left(2^{r}\cdot 2 \right) -3\cdot2^r -2r -3 \\ &=& 2^{r}\left( 6-3\right) -2r -3 \\ \mathbf{ s_{1\ldots r} } &=& \mathbf{ 3\cdot 2^{r} -2r -3 } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{ s_{1\ldots r} } &=& \mathbf{ 3\cdot 2^{r} -2r -3 } \quad | \quad r = 20 \\\\ s_{1\ldots 20} &=& 3\cdot 2^{20} -2\cdot(20) -3 \\ &=& 3\cdot 2^{20} -2\cdot(20) -3 \\ &=& 3\cdot 2^{20} -43 \\ &=& 3\cdot 1048576 -43 \\ &=& 3\cdot 1048576 -43 \\ &=& 3145728 -43 \\ \mathbf{ s_{1\ldots 20} } &=& \mathbf{ 3145685} \\ \hline \end{array}\)

 


laugh

 Nov 3, 2019

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