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Another math question

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Guest Feb 20, 2015

#2
+92744
+5

Here's another way to see this.......

The cumulative sum of the total entries of n rows is given by.....n( n + 1)

And we can always figure out the row of the entry we're looking for by taking the ceiling function answer of the positive root of n^2 + n - k = 0   where k = the nth entry .... so

$${{\mathtt{n}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{n}}{\mathtt{\,-\,}}{\mathtt{40}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{n}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{161}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\ {\mathtt{n}} = {\frac{\left({\sqrt{{\mathtt{161}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\ \end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{n}} = -{\mathtt{6.844\: \!288\: \!770\: \!224\: \!760\: \!2}}\\ {\mathtt{n}} = {\mathtt{5.844\: \!288\: \!770\: \!224\: \!760\: \!2}}\\ \end{array} \right\}$$

So ceiling(5.8)  = 6  ...so the 40th entry will be in row 6

And the value of each entry on row n = 2n......so....the 40th entry will be 2(6) = 12

CPhill  Feb 20, 2015
#1
+17746
+5

Row 1 has 2 numbers; row 2 has 4 numbers; row 3 has 6 numbers; row 4 has 8 numbers; row 5 has 10 numbers.

So far, we have 2 + 4 + 6 + 8 + 10 = 30 numbers. The next row has 12 numbers, so this row (the 6th row) contains the 40th number.

Row 1 contain 2's, row 2 contains 4's, row 3 contains 6's, row 4 contains 8's, row 5 contains 10's, and row 6 contains 12's.

geno3141  Feb 20, 2015
#2
+92744
+5

Here's another way to see this.......

The cumulative sum of the total entries of n rows is given by.....n( n + 1)

And we can always figure out the row of the entry we're looking for by taking the ceiling function answer of the positive root of n^2 + n - k = 0   where k = the nth entry .... so

$${{\mathtt{n}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{n}}{\mathtt{\,-\,}}{\mathtt{40}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{n}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{161}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\ {\mathtt{n}} = {\frac{\left({\sqrt{{\mathtt{161}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\ \end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{n}} = -{\mathtt{6.844\: \!288\: \!770\: \!224\: \!760\: \!2}}\\ {\mathtt{n}} = {\mathtt{5.844\: \!288\: \!770\: \!224\: \!760\: \!2}}\\ \end{array} \right\}$$

So ceiling(5.8)  = 6  ...so the 40th entry will be in row 6

And the value of each entry on row n = 2n......so....the 40th entry will be 2(6) = 12

CPhill  Feb 20, 2015