Question 1205092
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Tutor mccravyedwin has covered problem 1. 
I'll take a look at problem 2.


{{{(a+b)^n = sum(nCk*a^(n-k)*b^k,k=0,n)}}} The binomial theorem where n is an integer and {{{n >= 0}}}


{{{(3+(-2))^n = sum(nCk*3^(n-k)*(-2)^k,k=0,n)}}} Plug in a = 3 and b = -2


{{{(3-2)^n = sum(nCk*3^(n-k)*(-1*2)^k,k=0,n)}}}


{{{(1)^n = sum(nCk*3^(n-k)*(-1)^k*2^k,k=0,n)}}}


{{{1 = sum((-1)^k*nCk*3^(n-k)*2^k,k=0,n)}}}


{{{sum((-1)^k*nCk*3^(n-k)*2^k,k=0,n) = 1}}} where {{{n >= 0}}}




The nCk refers to the combination formula
{{{nCk = (n!)/(k!*(n-k)!)}}}
The nCk values are found in Pascal's Triangle.
The nCk replaces the notation (n¦k) which is often written as {{{(matrix(2,1,n,k))}}}
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