Interior numbers begin in the third row of Pascal's Triangle. The sum of the interior numbers in the fourth row is 6. The sum of the interior numbers of the fifth row is 14. What is the sum of the interior numbers of the seventh row?

Guest Nov 29, 2017

#1**+1 **

I noticed a pattern a long time ago...

The direct numbers can be found by 11^{r}, and the sum can be found using 2^{r}. This means that the sum of the interior numbers on row 7 = 2^{7} - 2, or 126. As a matter of fact, the function is: \(\sum\triangle(r) = maximum([0, 2^{r}-2])\), where maximum (array) = the maximum value in the array.

helperid1839321
Nov 29, 2017

#3**+1 **

You're right CPhill! I made a dumb mistake. Thanks for checking my answer!

helperid1839321
Nov 29, 2017

#2**+1 **

The sum of the interior integers in the nth row of Pascal's Triangle in your scheme is :

2^{n -1} - 2 [ where n is an integer > 2 ]

So....the sum of the interior intergers in the 7th row is 2^{(7-1)} - 2 = 2^{6} - 2 = 62

CPhill
Nov 29, 2017