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1. Find the coefficient of \(x^3y^3z^2\) in the expansion of \((x+y+z)^8\).

 

2. For each integer \(n\), let \(f(n)\) be the sum of the elements of the \(n\) th row (i.e. the row with \(n+1\) elements) of Pascal's triangle minus the sum of all the elements from previous rows. For example, \(f(2) = \underbrace{(1 + 2 + 1)}_{\text{2nd row}} - \underbrace{(1 + 1 + 1)}_{\text{0th and 1st rows}} = 1. \) What is the minimum value of \(f(n)\) for \(n \ge 2015\)?

tertre  Mar 8, 2018
 #1
avatar+91027 
0

Here's the first, tertre

 

( x + y + z)^8  =   ( x  +  ( y + z) )^8  = 

 

C (8,5) * x^3 * (y + z)^5   .......we need to  find the coefficiient on y^3z^2  for the second part in red        

 

C (8,5) * x^3 * C(5,3)* y^3 * z^2  =

 

56* x^3  *  10 * y^3 * z^2   =

 

560   (x^3 * y^3 * z^2)

 

 

cool cool cool

CPhill  Mar 8, 2018
edited by CPhill  Mar 8, 2018
 #2
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0

Just a minor typo in the 3rd to the last step. The second x^3 should read y^3.

Guest Mar 8, 2018
 #3
avatar+144 
+2

It will always be \(2^n-(2^{n-1}+2^{n-2} \cdots +2^1+2^0)=1\).Thus, the minimum value is \(\boxed{1}\)

azsun  Mar 8, 2018

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