Consider the expansion of (x+3)^10. Find the term containing x^3.

Question

please help. Calculus!

 

Answer 1

Use Binomial Theorem:

 

(a+b)^n= Σ (I= 0 to n) C(n,i)a^i*b^(n-i)

 

The term you want is C(3,10)*x^3*3^7  where a=x, b=3, i=3 and n=10

 

10!/(3!*7!)*3^7*x^3=240*3^7*x^3

 

Answer 2

(A + B)ⁿ = _nC₀A^n-0B⁰ + _nC₁A^n-1B¹ + _nC₂A^n-2B² + ... +_nC_nA^n-nBⁿ

 

(x + 3)¹⁰ = ₁₀C₀x¹⁰3⁰ + ₁₀C₁x⁹3¹ + ₁₀C₂x⁸3² + ... + ₁₀C₁₀x⁰3¹⁰

 

The term involving x³ is the 8th term and is equal to: 

 

        ₁₀C₇x³3⁷ =  262440x³

 

                                                                                         

 

 

Answer 3

To expand a binomial like this, use the binomial coefficients from Pascal's Triangle. We're interested in the 10^th row, since your binomial is raised to the 10^th power:

 

0:  1

1:  1 1

2:  1 2 1

3:  1 3 3 1

4:  1 4 6 4 1

5:  1 5 10 10 5 1

6:  1 6 15 20 15 6 1

7:  1 7 21 35 35 21 7 1

8:  1 8 28 56 70 56 28 8 1

9:  1 9 36 84 126 126 84 36 9 1

10: 1 10 45 120 210 252 210 120 45 10 1

 

Our binomial will be expanded like this, where a_i represents the i^th coefficient in Pascal's Triangle's 10^th row (a₁ = 1):

 

x¹⁰ + a₂x⁹3 + a₃x⁸3² + a₄x⁷3³ + a₅x⁶3⁴ + a₆x⁵3⁵ + a₇x⁴3⁶ + a₈x³3⁷ + a₉x²3⁸ + a₁₀x3⁹ + 3¹⁰

 

For each successive term, we advance by one number in the triangle, x's exponent goes down by 1, and 3's exponent goes up by 1.

 

The x³ term is:

 

a₈x³3⁷ = 120x³2187 = 262440x³