Please check this for me
expand (3z-2b)^5 using Binomial Theorem
(a+b)^n = n on top sigma in middle k=0 to right (n/k)a^(n-k)b^k
(3a-2b)^5 = 5 on top sigma in middle k=0 on bottom(5/k)(3a)^5-k(-2b)^k
(5/0)(3a)^5(-2b)^0+(5/1)(3a)^4(-2b)^1 +(5/2)(3a)^3(-2b)^2
+(5/3)(3a)^2(-2b)^3 + (5/4)(3a)^1(-2b)^4+(5/5)(2a)^0(-2b)^5
=243a^5-810a^4b+1080a^3b^2+-720a^2b^3+240ab^4-32b^5
5!/(5-5)!5!=5!/1-5!=5!/5!1
(5/0)=5!/(5-0)!0!=5!/5!0!=1
please delete last two lines-that was just my work I calculated- it shouldn't e there
actual answer is 243a^5-810a^4b + 1080a^3b^2 -720a^2 b^3 + 240ab^4-32b^5
Don't know if you have to go into sigma notation unless your instructor wants you to do it that way.
I would have in front of me Pascal's Triangle , where I would need the row
1 5 10 10 5 1
so (3a - 2b)^5
= 1(3a)^5 + 5(3a)^4 (-2b) + 10(3a)^3 (-2b)^2 + 10(3a)^2 (-2b)^3 + 5(3a) (-2b)^4 +1(-2b)^5
= 243a^5 - 810a^4 b + 1080a^3 b^2 - 720a^2 b^3 + 240a b^4 - 32b^5
Your answer is correct
Thank you for checking
To expand (3z-2b)^5 using the Binomial Theorem, you can follow these steps:
Step 1: Write down the general form of the Binomial Theorem:
(a+b)^n = Σ(k=0 to n) (n/k)a^(n-k)b^k
Step 2: Plug in the values from your expression:
(3z-2b)^5 = Σ(k=0 to 5) (5/k)(3z)^(5-k)(-2b)^k
Step 3: Simplify each term using the values of k and the exponents:
(5/0)(3z)^5(-2b)^0 + (5/1)(3z)^4(-2b)^1 + (5/2)(3z)^3(-2b)^2 + (5/3)(3z)^2(-2b)^3 + (5/4)(3z)^1(-2b)^4 + (5/5)(3z)^0(-2b)^5
Step 4: Calculate each term:
(5/0)(3z)^5(-2b)^0 = 1(3z)^5 = 243z^5
(5/1)(3z)^4(-2b)^1 = 5(3z)^4(-2b) = -810z^4b
(5/2)(3z)^3(-2b)^2 = 10(3z)^3(-2b)^2 = 1080z^3b^2
(5/3)(3z)^2(-2b)^3 = 10(3z)^2(-2b)^3 = -720z^2b^3
(5/4)(3z)^1(-2b)^4 = 5(3z)^1(-2b)^4 = 240zb^4
(5/5)(3z)^0(-2b)^5 = 1(3z)^0(-2b)^5 = -32b^5
Step 5: Combine all the terms obtained:
243z^5 - 810z^4b + 1080z^3b^2 - 720z^2b^3 + 240zb^4 - 32b^5
So, expanding (3z-2b)^5 using the Binomial Theorem gives us the result: 243z^5 - 810z^4b + 1080z^3b^2 - 720z^2b^3 + 240zb^4 - 32b^5.