The coefficients for the binomial expansion of (a + b)³ are 1, 3, 3, 1.
We have,
The binomial expansion (a + b)³,
So,
Now,
We know that,
Formula for binomial expansion,
i.e.
(u + v)ⁿ = ⁿC₀ · uⁿ · v⁰ + ⁿC₁ · uⁿ⁻¹ · v¹ + ⁿC₂ · uⁿ⁻² · v² + ⁿC₃ · uⁿ⁻³ · v³
Now,
According to the question,
Here,
u = a,
v = b,
And, n =3
Now,
Putting values,
i.e.
(a + b)³ = ³C₀ · a³ · b⁰ + ³C₁ · a³⁻¹ · b¹ + ³C₂ · a³⁻² · b² + ³C₃ · a³⁻³ · b³
We get,
(a + b)³ = ³C₀ · a³ · b⁰ + ³C₁ · a² · b¹ + ³C₂ · a¹ · b² + ³C₃ · a⁰ · b³
Now,
Using combination formula,
[tex]^{n} C_{r} = \frac{n!}{r!(n-r)!}[/tex]
On solving we get,
(a + b)³ = a³ + 3 a² b + 3 a b² + b³
So,
Now,
The coefficients of :
a³ = 1,
3a²b = 3,
3ab² = 3,
b³ = 1,
So,
The coefficients of the given binomial expansion are 1, 3, 3, 1.
Hence we can say that the coefficients for the binomial expansion of (a + b)³ are 1, 3, 3, 1.
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