+0

+1
294
4

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
+555
+1

I noticed a pattern a long time ago...

The direct numbers can be found by 11r, and the sum can be found using 2r. This means that the sum of the interior numbers on row 7 = 27 - 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
+555
+1

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

helperid1839321  Nov 29, 2017
#2
+88898
+1

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

2n -1  -  2       [ where n is an integer > 2 ]

So....the sum of the interior intergers in the 7th row is   2(7-1) - 2  = 26 - 2  = 62

CPhill  Nov 29, 2017
#4
+88898
+1

Actually, helperid, your answer is probably more correct......we usually number the first row as "row 0 "

So...the first row where we will have an interior integer is row 2

So....the answer of 2n - 2   for n ≥ 2 is actually better given the traditional numbering scheme

CPhill  Nov 29, 2017