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