Binomial Expansion

[YJ's]

Please help me with Binomial Expansion...

 

I really don't understand how to solve this... but I need to understand it quickly T_T

 

Please solve this question with clear explanation for every step... Thank you so much!

Edited November 6, 2016 by YJ's

[kw0573]

I'll do #2 as an example. Find coefficient of x³ in expansion of (2x² - 3/x)⁶. This is just a condensed explanation, if you have trouble understanding any step please ask!
1. Locate some exponent r such that (x²)^6-r * (x^-1)^r = x³. You can solve 2(6 - r) + (-1)r = 3 to get r = 3.
2. Use Data Booklet equation found under 1.3, which says the general term is (n choose r) * a^n-rb^r. In our question n = 6, r = 3, a = 2x², b = (-3/x)

3. (6 choose 3) = 20.

a^n-r = (2x²)³ = 8x⁶.

b^r = (-3/x)³ = -27 x^-3

4. Multiply all together: 20 * 8x⁶ * (-27 x^-3) = -4320 x³. Coefficient of x³ term is -4320. 

 

Edited November 6, 2016 by kw0573

[kw0573]

Guess I'll also do number 5. Find coefficient of x⁶ in (2 - x)¹(3x + 1)⁹
1. Similar step, but now we use variables r and s instead of just r in question 2. (x⁰)^{1 - r}(x)^r (x)^{9 - s} (x⁰)^s = x⁶. 

r + (9 - s) = 6, r - s = -3. Because r can only take on values 0 or 1. If r = 0, s = 3; if r = 1, s = 4. 

2. for r = 0, s = 3, we get the term (1 choose 0) * (2)^1-0 (-x)⁰ * (9 choose 3) * (3x)^9-3 * (1)³ = 1 * 2 * 1 * 84 * 729x⁶ * 1 = 122472x⁶

....for r = 1, s = 4, we get the term (1 choose 1) * (2)^1-1 (-x)¹ * (9 choose 4) * (3x)^9-4 * (1)⁴ = 1 * 1 * (-x) * 126 * 243x⁵ * 1 = -30618x⁶

3. Add them up: (122472 - 30618) x⁶ = 91854 x⁶. The coefficient of x⁶ term is 91854.
 

Edited November 6, 2016 by kw0573