How do you use the Binomial theorem to expand #(a-3b)^5#?

1 Answer
Binayaka C.
Jan 12, 2018

#(a -3 b)^5 =a^5-15 a^4b+90 a^3b^2-270a^2b^3+405 ab^4-243b^5 #

Explanation:

Binomial theorem:

#(a + b)^n =(nC_0) a^nb^0+ (nC_1) a^(n-1)b^1 +(nC_2) a^(n-2)b^2 +...... (nC_n)b^n#

Here # (a=3b ; b=-3b , n= 5) #

We know #nC_r= (n!)/(r!(n-r)!) :. 5C_0=1 , 5C_1=5 , 5C_2=10 ,5C_3=10, 5C_4=5, 5C_5=1#

#(a -3 b)^5 =(5C_0) a^5(-3b)^0+ (5C_1) a^4(-3b)^1 +(5C_2) a^3(-3b)^2 +(5C_3) a^2(-3b)^3+(5C_4) a^1(-3b)^4+(5C_5) a^0(-3b)^5 # or

#(a -3 b)^5 =a^5-15 a^4b+90 a^3b^2-270a^2b^3+405 ab^4-243b^5 # [Ans]

Related questions
See all questions in Pascal's Triangle and Binomial Expansion
Impact of this question
4529 views around the world
You can reuse this answer
Creative Commons License