what is the 5th term of (2x-3y)^10

Question

campbell_st · Answer

well it requires binomial expansion and combinations 

for (a + b)^n the expansion is 

$$^nC_{r} a^{n - r}b^r$$

in your question n = 10, r = 4 a = 2x and b = -3y

just substitute and evaluate. 

the reason that r = 4 is because the 1st term has  r = 0

anonymous · Answer

can you teach me step by step  for my final exam tommorow

campbell_st · Answer

well the expansion is as above 

the 1st and last terms have coefficients of 1

the 2nd and 2nd last terms have coefficients of the power

so the formula for above can be used for any binomial expansion 

the expansion starts 

$$^{10}C_{0}(2x)^{10}(-3y)^0 + ^{10}C_{1}(2x)^9(-3y)^1 + ^{10}C_{2}(2x)^8(-3y)^2+.... $$

which is 

$$1024x^{10} -15360x^9y + 103680x^8y^2-.... $$

thats all thats needed....

pixiedust1 · Answer

Woah... That's hard!! What grade math is this ?

campbell_st · Answer

well its year 11

anonymous · Answer

@campbell_st i dont get it

ranga · Answer

Substituting the numbers in campbell_st's first reply you get:$$\Large ^{10}C _{4}(2x)^6(-3y)^4$$Simplify.

ranga · Answer

$$\Large ^{n}C _{r} = \frac{ n! }{ r!(n-r)! }$$