Apply the binomial theorem to find the first three terms in ascending powers of x of (1 + x)ⁿ, where n=32
for any n, we have
(1+x)^n = 1^n + n*1^(n-1)x + n(n-1)/2 1^(n-2)x^2 + ...
= 1 + nx + n(n-1)/2 x^2 + ...
So, plug in n=32
To apply the binomial theorem, we use the formula:
(1 + x)ⁿ = C(n, 0)⋅x⁰ + C(n, 1)⋅x¹ + C(n, 2)⋅x² + ... + C(n, n-1)⋅xⁿ-¹ + C(n, n)⋅xⁿ
Here, C(n, r) represents the binomial coefficient, which is calculated as follows:
C(n, r) = n! / ((n-r)! ⋅ r!)
In this case, n = 32. Let's find the first three terms in ascending powers of x:
1. Term with x⁰:
Plugging in r = 0, we get:
C(32, 0)⋅x⁰ = 1⋅x⁰ = 1
Therefore, the first term is 1.
2. Term with x¹:
Plugging in r = 1, we get:
C(32, 1)⋅x¹ = (32! / (32-1)!⋅1!)⋅x¹ = 32⋅x¹ = 32x
Therefore, the second term is 32x.
3. Term with x²:
Plugging in r = 2, we get:
C(32, 2)⋅x² = (32! / (32-2)!⋅2!)⋅x² = (32⋅31 / 2!⋅1!)⋅x² = (32⋅31 / 2)⋅x² = 31⋅16x²
Therefore, the third term is 31⋅16x².
So, when n = 32, the first three terms of (1 + x)ⁿ in ascending powers of x are:
1 + 32x + 31⋅16x²
To find the first three terms in ascending powers of x of (1 + x)ⁿ, where n=32, we can use the binomial theorem. The binomial theorem states that for any positive integer n:
(1 + x)ⁿ = nC₀ * 1ⁿ * x⁰ + nC₁ * 1ⁿ⁻¹ * x¹ + nC₂ * 1ⁿ⁻² * x² + ... + nCₖ * 1ⁿ⁻ᵏ * xᵏ + ...
where nCᵢ represents the binomial coefficients, given by nCᵢ = n! / (i! * (n - i)!), and k ranges from 0 to n.
Let's calculate the first three terms using this formula. Given n=32, we have:
Term 1: nC₀ * 1ⁿ * x⁰ = 32C₀ * 1³² * x⁰ = 1 * 1³² * 1 = 1.
Term 2: nC₁ * 1ⁿ⁻¹ * x¹ = 32C₁ * 1³¹ * x = 32 * 1³¹ * x = 32x.
Term 3: nC₂ * 1ⁿ⁻² * x² = 32C₂ * 1³⁰ * x² = (32 * 31) / (2 * 1) * 1³⁰ * x² = 496 * x².
Therefore, the first three terms in ascending powers of x of (1 + x)³² are:
Term 1: 1
Term 2: 32x
Term 3: 496x².