Respuesta :
The answers are calculated according to the binomial theorem as follows:
(a) The expansion of (1 + x)² = 1 + 2x + x²
(b) The term in x⁴ in the expansion of (1 + cx)(1 + x)³ is 'c' and the expansion is: 1 + cx + 3x + 3cx² + 3x² + 3cx³ + 3x³ + cx⁴
(c) The possible value of c is 15. The expansion of (1 + cx)(1 + x)¹ is:
(1 + cx)(1 + x)¹ = 1 + (1 + c)x¹ + cx²
What is the binomial theorem for the integral index?
If n is a positive integer then
(x + a)ⁿ = ⁿC₀xⁿ + ⁿC₁xⁿ⁻¹a + ⁿC₂xⁿ⁻²a² + ... + [tex]^nC_rx^{n-r}a^r[/tex] + ... + ⁿCₙaⁿ
The expansion contains (n + 1) terms.
In the expansions, the coefficients ⁿC₀, ⁿC₁, ⁿC₂, ...,ⁿCₙ are called binomial coefficients.
And
(1 + x)ⁿ = ⁿC₀ + ⁿC₁x + ⁿC₂x² + ... + ⁿCₙxⁿ = C₀ + C₁x + C₂x² + ... + Cₙxⁿ
Calculation:
(a) Expansion of (1 + x)²:
The given integer index is (1 + x)²; where n = 2 and a = 1
So, according to the binomial theorem,
(1 + x)² = ²C₀ + ²C₁x + ²C₂x²
= 1 + 2x + x²
(Since we have [tex]^nC_r = \frac{n!}{r!(n-r)!}[/tex])
(b) Expansion of (1 + cx)(1 + x)³:
Here the index is (1 + x)³; n = 3
Then,
(1 + x)³ = ³C₀ + ³C₁x + ³C₂x² + ³C₃x³
= 1 + 3x + 3x² + x³
So,
(1 + cx)(1 + x)³ = (1 + cx)(1 + 3x + 3x² + x³)
⇒ 1 + 3x + 3x² + x³ + cx + 3cx² + 3cx³ + cx⁴
⇒ 1 + cx + 3x + 3cx² + 3x² + 3cx³ + x³ + cx⁴
From this expansion, the coefficient of x⁴ is 'c'. So, the x⁴ term is 'c'.
(c) Expansion of (1 + cx)(1+x)¹:
The index is (1 + x)¹; n = 1
So, it remains the same(there is no further expansion)
Then,
(1 + cx)(1+x)¹ = (1 + cx)(1 + x)
On simplifying,
(1 + cx)(1+x)¹ = 1 + x + cx + cx²
Since it is given that, the expansion includes 16x¹, we can write
x + cx = 16x
⇒ (1 + c) = 16
⇒ c = 16 - 1
∴ c = 15
Learn more about the Binomial expnsion here:
https://brainly.com/question/13602562
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