The expansion of a binomial is given by :

##(x+y)^n=( (n) (0) )*x^n+( (n) (1) )*x^(n-1)*y^1+…+( (n) (k) )*x^(n-k)*y^k+…+( (n) (n) )*y^n = sum_(k=0)^n*( (n) (k) )*x^(n-k)*y^k ##

where ##x y in RR## ##k n in NN## and ##( (n) (k) )## denotes combinations of ##n## things taken ##k## at a time.

## ( (n) (k) )*x^(n-k)*y^k ## is the general term of the binomial expansion.

We also have the formula: ##( (n) (k) )=(n!)/(k!*(n-k)!)## where ##k! = 1*2*…*k##

We have three cases:

Case 1: If the terms of the binomial are a variable and a constant ##(y=c## where ##c## is a constant) we have ##(x+c)^n=( (n) (0) )*x^n+( (n) (1) )*x^(n-1)*c^1+…+( (n) (k) )*x^(n-k)*c^k+…+( (n) (n) )*c^n ##

We can see that the constant term is the last one: ##( (n) (n) )*c^n##

(as ##( (n) (n) )## and ##c^n## are constant their product is also a constant).

Case 2: If the terms of the binomial are a variable and a ratio of that variable (##y=c/x## where ##c## is a constant) we have:

## (x+c/x)^n=( (n) (0) )*x^n + ( (n) (1) )*x^(n-1)*(c/x)^1+…+( (n) (k) )*x^(n-k)*(c/x)^k+…+( (n) (n) )*(c/x)^n ##

This time we see that the constant term is not to be found at the extremities of the binomial expansion. So we should have a look at the general term and try to find out when it becomes a constant:

## ( (n) (k) )*x^(n-k)*(c/x)^k=( (n) (k) )*x^(n-k)*c^k*1/x^k = (( (n) (n) )*c^k)*(x^(n-k))/x^k = (( (n) (k) )*c^k)*x^(n-2k) ##.

We can see that the general term becomes constant when the exponent of variable ##x## is ##0##. Therefore the condition for the constant term is: ##n-2k=0 rArr## ##k=n/2## . In other words in this case the constant term is the middle one (##k=n/2##).

Case 3: If the terms of the binomial are two distinct variables ##x## and ##y## such that ##y## cannot be expressed as a ratio of ##x## then there is no constant term . This is the general case ##(x+y)^n##

The post How do I find the constant term of a binomial expansion? appeared first on homeworkcrew.

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