Binomial Expansion of (1+x)^n: Coefficient of x^n

[sara_87]

Apply the binomial expansion to : (1+x)^n and show that the coefiicient of x^n in the expansion of (1+x)^2n is:
(nC0)^2 +(nC1)^2 +...+(nCn)^2
hint: (nCm)=(nC(n-m))

my approach:

(1+x)^n = x^n + nx^(n-1) + (nC2)x^(n-2) +...+ 1

(1+x)^2n = x^(2n) + nx^(2n-1) +...+ x^n

i don't know what to do next. it looks easy but i can't figure it out.
can someone help me please?.

 

[EnumaElish]

(1+x)^2n = ((1+x)^n)^2 = (x^n + nx^(n-1) + (nC2)x^(n-2) +...+ 1)(x^n + nx^(n-1) + (nC2)x^(n-2) +...+ 1).

Now you need to cross-multiply and verify which cross-multiplied terms simplify to x^n. For ex., (Ax)Bx^(n-1) = (AB)x^n.

 

[tacman]

Sure, I can help you with this problem. Let's start by reviewing the binomial expansion formula:

(a+b)^n = (nC0)a^n + (nC1)a^(n-1)b + (nC2)a^(n-2)b^2 + ... + (nCn)b^n

Using this formula, we can expand (1+x)^2n as:

(1+x)^2n = (2nC0)1^(2n) + (2nC1)1^(2n-1)x + (2nC2)1^(2n-2)x^2 + ... + (2nC2n)x^(2n)

Notice that all the terms with powers greater than n will have a coefficient of 0 since they involve x raised to a power greater than n. Also, the terms with even powers of x will have coefficients equal to the corresponding (2nCm) terms, while the terms with odd powers of x will have coefficients equal to (2nCm) multiplied by the power of x.

So, the coefficient of x^n in the expansion of (1+x)^2n will be the sum of all the coefficients of x^n in the expanded terms. This can be written as:

Coefficient of x^n = (2nC0) + (2nC2)x^n + (2nC4)x^(2n-2) + ... + (2nC2n)x^0

Now, let's use the hint given in the problem: (nCm) = (nC(n-m)). This means that the coefficient of x^n in the expanded term (2nCm)x^(2n-2m) will be equal to the coefficient of x^m in the expanded term (2nC(n-m))x^(2n-2(n-m)). In other words, the coefficient of x^n in the expanded term (2nCm)x^(2n-2m) is equal to the coefficient of x^(n-m) in the expanded term (2nC(n-m))x^(n).

Applying this to our original equation, we can rewrite the coefficient of x^n as:

Coefficient of x^n = (2nC0) + (2nC2)x^n + (2nC4)x^(2n-2) + ... + (2nC2n)x^0