Posted by Bethany on Wednesday, January 4, 2012 at 12:54pm.
Please check this for me
expand (3z2b)^5 using Binomial Theorem
(a+b)^n = n on top sigma in middle k=0 to right (n/k)a^(nk)b^k
(3a2b)^5 = 5 on top sigma in middle k=0 on bottom(5/k)(3a)^5k(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^5810a^4b+1080a^3b^2+720a^2b^3+240ab^432b^5
5!/(55)!5!=5!/15!=5!/5!1
(5/0)=5!/(50)!0!=5!/5!0!=1

PreCalcplease check  Bethany, Wednesday, January 4, 2012 at 1:03pm
please delete last two linesthat was just my work I calculated it shouldn't e there
actual answer is 243a^5810a^4b + 1080a^3b^2 720a^2 b^3 + 240ab^432b^5

PreCalcplease check  Reiny, Wednesday, January 4, 2012 at 1:09pm
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

PreCalcplease check  Bethany, Wednesday, January 4, 2012 at 1:16pm
Thank you for checking
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