SOLUTION: Hello, I'm at a loss of how to solve this. Any help would be very much appreciated. Thanks in advance! In Pascal's triangle, nCr+nCr+1=n+1Cr+1. Let X, Y, and Z be three consecut

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Question 1083531: Hello, I'm at a loss of how to solve this. Any help would be very much appreciated. Thanks in advance!
In Pascal's triangle, nCr+nCr+1=n+1Cr+1. Let X, Y, and Z be three consecutive numbers in the nth row of Pascal's Triangle, such that X = nCr, Y = nCr+1, and Z = nCr+2. Use the nature of Pascal's Triangle to show that X + Y + Z = n+2Cr+2 - nCr+1.
Hint: Start by finding X + Y and Y + Z.
Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
To keep from getting bogged down with the n's and r's
Let's write the "Pascal's triangle formula" using
letters different from n and r:

sCt + sC(t+1) = (s+1)C(t+1)  <-- Pascal's triangle formula 

X = nCr, Y = nCr+1, and Z = nCr+2

X + Y = nCr + nC(r+1) = (n+1)C(r+1)

Y + Z = nC(r+1) + nC(r+2) 

Let s = n and t = r+1 in Pascal's triangle formula:

Y+Z = nC(r+1) + nC(r+1+1) = nC(r+1) + nC(r+2)

Adding the two equations together:

X + 2Y + Z = (n+1)C(r+1) + (n+1)C(r+2),

Now let s = n+1, and t = r+1 in Pascal's triangle formula:

(n+1)C(r+1) + (n+1)C(r+1+1) = (n+1+1)C(r+1+1) 

which simplifies to

(n+1)C(r+1) + (n+1)C(r+2) = (n+2)C(r+2) 

That left side turned out to be the same as the right side of 
X + 2Y + Z above. So

X + 2Y + Z = (n+1)C(r+1) + (n+1)C(r+2) = (n+2)C(r+2)

or

X + 2Y + Z = (n+2)C(r+2)


To get X + Y + Z we need to subtract Y = nC(r+1) from
both sides:

X + 2Y + Z = (n+2)C(r+2)
     Y     =   nC(r+1)
------------------------
X  + Y + Z = (n+2)C(r+2) - nC(r+1)

Edwin
 

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