# Formula for the n-th row of Pascal's Triangle

1. Jan 22, 2013

### stgermaine

1. The problem statement, all variables and given/known data
Find a formula for the sum of the elements of the nth row of Pascal's Triangle

2. Relevant equations
C(n,r) = [C(n-1,r-1) + C(n-1,r)]
C(n,0) = C(n,n) = 1

3. The attempt at a solution

I started with the summation of the elements in the rows n
$\sum^{n}_{r=0} C(n,r)$
= $C(n,0) + \sum^{n-1}_{r=1} C(n,r) + C(n,n)$
= $C(n,0) + \sum^{n-1}_{r=1} [C(n-1,r-1) + C(n-1,r)] + C(n,n)$
= $C(n,0) + \sum^{n-1}_{r=1} C(n-1,r-1) + \sum^{n-1}_{r=1} C(n-1,r) + C(n,n)$
= $C(n,0) + \sum^{n-2}_{r=0} C(n-1,r) + \sum^{n-1}_{r=1} C(n-1,r) + C(n,n)$
= $C(n,0) + [C(n-1,0) + \sum^{n-2}_{r=1} C(n-1,r)] + [\sum^{n-2}_{r=1} C(n-1,r) + C(n-1,n-1)] + C(n,n)$

= $2 \cdot C(n-1,0) + \sum^{n-2}_{r=1} C(n-1,r) + \sum^{n-2}_{r=1} C(n-1,r) + 2 \cdot C(n-1,n-1)$ since C(n,0) = 1 = C(n-1,0) and C(n,n) = 1 = C(n-1,n-1)
=$2 \cdot \sum^{n-1}_{r=0} C(n-1,r)$

This gives the intuitively correct response that the sum of the nth row is twice that of the (n-1)th row. How can I go from here to sum = 2^n?

Thank you!

2. Jan 22, 2013

### jbunniii

Do you know the binomial theorem? What is a formula for $(x+y)^n$ in terms of $C(n,r)$?

3. Jan 22, 2013

### jbunniii

This should be a simple induction. What is the sum of the 1st row?