How do I find the constant term of a binomial expansion?

2 Answers
Sep 26, 2014

The constant term of (x+a)^n(x+a)n is always a^nan; for example, the constant term of (x+3)^7(x+3)7 is 3^737.

Jun 3, 2015

The expansion of a binomial is given by the Binomial Theorem:

(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

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