+0

# Another math question

0
242
2

Guest Feb 20, 2015

#2
+80913
+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
Sort:

#1
+17711
+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
+80913
+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

### 26 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details