Binomial Expansion of (1 + x + x^2)^n
The binomial theorem provides a formula to expand expressions of the form (a + b)^n, where n is a positive integer. However, it doesn't directly apply to expressions like (1 + x + x^2)^n. To handle this, we need to employ a clever trick: re-write the expression as a product of two binomials.
Step 1: Re-write the expression
We can re-write (1 + x + x^2)^n as [(1 + x) + x^2]^n. Now we have an expression in the form (a + b)^n, where a = (1 + x) and b = x^2.
Step 2: Apply the binomial theorem
Using the binomial theorem, we can expand [(1 + x) + x^2]^n as:
(1 + x + x^2)^n = [(1 + x) + x^2]^n = ∑(k=0 to n) (n choose k) (1 + x)^(n-k) (x^2)^k
where (n choose k) represents the binomial coefficient, calculated as n!/(k!(n-k)!).
Step 3: Expand (1 + x)^(n-k)
We can now expand (1 + x)^(n-k) using the binomial theorem again:
(1 + x)^(n-k) = ∑(j=0 to n-k) (n-k choose j) (1)^(n-k-j) (x)^j
Step 4: Combine the expansions
Substituting this back into our previous equation and simplifying:
(1 + x + x^2)^n = ∑(k=0 to n) ∑(j=0 to n-k) (n choose k) (n-k choose j) x^(j+2k)
Final Expression
The binomial expansion of (1 + x + x^2)^n is given by:
(1 + x + x^2)^n = ∑(k=0 to n) ∑(j=0 to n-k) (n choose k) (n-k choose j) x^(j+2k)
This expression provides a general formula for the expansion, where the coefficients are determined by the binomial coefficients. It's important to note that this expansion will contain terms of x raised to powers ranging from 0 to 2n.
Example:
Let's find the binomial expansion of (1 + x + x^2)^2:
(1 + x + x^2)^2 = ∑(k=0 to 2) ∑(j=0 to 2-k) (2 choose k) (2-k choose j) x^(j+2k)
For k = 0:
∑(j=0 to 2) (2 choose 0) (2 choose j) x^(j) = 1 + 2x + x^2
For k = 1:
∑(j=0 to 1) (2 choose 1) (1 choose j) x^(j+2) = 2x^2 + 2x^3
For k = 2:
∑(j=0 to 0) (2 choose 2) (0 choose j) x^(j+4) = x^4
Adding these terms together:
(1 + x + x^2)^2 = 1 + 2x + 3x^2 + 2x^3 + x^4
This method allows us to systematically expand expressions like (1 + x + x^2)^n and obtain a general form for the result.
