Question

How do I do number 3

Answer 1

Adding up the sums of the 6 rows, we get
1
2
4
8
16
32

As you may be able to tell, all numbers here multiply by 2 to get to the next number - 1*2=2, 2*2=4, etc.. That means that this is a geometric sequence, or the first term (t₁) multiplied by the ratio between each number (2 in our case) to the power of (n-1), with n being the number in the sequence that we want (7 in this case). Thus, we have 1*(2⁷⁻¹)=2⁶=64 as the sum of the terms in the seventh sequence. Testing this out, we find 7 choose 0, 7 choose 1, etc. - I'll find the first and second to last coefficients using the formula

`\left[\begin{array}{ccc}x\\y\end{array}\right] = (x!)/((x-y)!y!)`

`\left[\begin{array}{ccc}7\\0\end{array}\right] = (7!)/((7-0)!0!) = (7*6*5*4*3*2*1)/(7*6*5*4*3*2*1*1) =1`

`\left[\begin{array}{ccc}7\\6\end{array}\right] = (7!)/((7-6)!6!) = (7*6*5*4*3*2*1)/(6*5*4*3*2*1*1) =7`

I challenge you to do the rest (if you don't already have the triangle)!
Our coefficients in the 7th row are as follows:
1 6 15 20 15 6 1
Adding them up, we get 64, so our conjecture was right!

Feel free to ask further questions!