If A be the sum of the odd terms and B the sum of the even terms in the expansion of
(x+a)ⁿ. Prove that A² - B² = (x²-a²)ⁿ.
To prove that A² - B² = (x² - a²)ⁿ, we need to find the formulas for A and B in terms of (x+a)ⁿ and then substitute them into the left-hand side of the equation.
We start by expanding (x+a)ⁿ using the Binomial Theorem, which states that:
(x+a)ⁿ = C(n,0) * xⁿ * a⁰ + C(n,1) * xⁿ⁻¹ * a¹ + C(n,2) * xⁿ⁻² * a² + ... + C(n,n-1) * x¹ * aⁿ⁻¹ + C(n,n) * x⁰ * aⁿ,
where C(n,r) is the binomial coefficient, given by C(n,r) = n! / (r! * (n-r)!).
Now let's group the terms in this expansion based on their parity.
Odd terms have an even exponent for x and an odd exponent for a. Even terms have an odd exponent for x and an even exponent for a.
For odd terms:
A = C(n,1) * xⁿ⁻¹ * a¹ + C(n,3) * xⁿ⁻³ * a³ + ... + C(n,n-1) * x¹ * aⁿ⁻¹.
For even terms:
B = C(n,0) * xⁿ * a⁰ + C(n,2) * xⁿ⁻² * a² + ... + C(n,n-2) * x² * aⁿ⁻² + C(n,n) * x⁰ * aⁿ.
Now let's simplify and rearrange these expressions.
A = a * (C(n,1) * xⁿ⁻¹ * a⁰ + C(n,3) * xⁿ⁻³ * a² + ... + C(n,n-1) * x¹ * aⁿ⁻²).
= a * (C(n,1) * xⁿ⁻¹ * a⁰ + C(n,3) * xⁿ⁻³ * a² + ... + C(n,n-1) * x¹ * aⁿ⁻² + C(n,n) * x⁰ * aⁿ * 0ⁿ⁻¹)
= a * (x + a)ⁿ - aⁿ.
= (x + a)ⁿ - aⁿ.
B = x * (C(n,0) * xⁿ⁻¹ * a⁰ + C(n,2) * xⁿ⁻³ * a² + ... + C(n,n-2) * x² * aⁿ⁻² + C(n,n) * x⁰ * aⁿ).
= x * (C(n,0) * xⁿ⁻¹ * a⁰ + C(n,2) * xⁿ⁻³ * a² + ... + C(n,n-2) * x² * aⁿ⁻² + C(n,n) * x⁰ * aⁿ + C(n,n) * x⁰ * aⁿ⁻²).
= x * (x + a)ⁿ - xⁿ.
= (x + a)ⁿ - xⁿ.
Now we can substitute these expressions into A² - B².
A² - B² = [(x + a)ⁿ - aⁿ]² - [(x + a)ⁿ - xⁿ]²
= [(x + a)ⁿ - aⁿ]² - [(x + a)ⁿ - xⁿ][(x + a)ⁿ - xⁿ]
= [(x + a)ⁿ - aⁿ][(x + a)ⁿ - aⁿ] - [(x + a)ⁿ - xⁿ][(x + a)ⁿ - xⁿ]
= [(x + a)ⁿ - aⁿ][(x + a)ⁿ - aⁿ - (x + a)ⁿ + xⁿ]
= [(x + a)ⁿ - aⁿ][xⁿ - aⁿ]
= (xⁿ + aⁿ - aⁿ)(xⁿ - aⁿ)
= (xⁿ - aⁿ)²
= (x² - a²)ⁿ.
Therefore, A² - B² = (x² - a²)ⁿ, as desired.