What is the third term in the expansion (x+2y)^5?

Question

How do I solve this, and what is the answer?

Answer 1

x^5+5x^4(2y)+10x^3(2y)^2
The third term is 10x^3(4y^2)=40x^3y^2

Answer 2

a million- the 1st 3 words, whilst accelerated, have here powers: First term: (x^2)^20 2d term: -A[(x^2)^19]x third term: B[(x^2)^18]x^2 the place A & B are unknown coefficients. Now, all of us understand the flexibility of x interior the third term is 38 2- to understand the coefficient of the third term (B), purely look at the triangle of coefficients primary as Pascal triangle. There you will discover the coefficient is a hundred ninety. pondering (a million) & (2) 190x^38 is the terrific answer. elect (a)!

Answer 3

It would be 10x³(2y)² = 40x³ y².

Answer 4

You need Pascal's triangle. It's on Wikipedia.

Answer 5

you're looking for

(5,3) x^2*(2y)^3

(5,3) = n!/k!(n-k)!

= 5!/(3!(2)!)

= 120 / 12

= 10

Therefore, the third term is

80x^2y^3

Answer 6

5n2 * (x+2y)^3 * 5^2

Answer 7

set u = (x+2y)^2
= x^2+4xy+4y^2
u^2 = (x^2+4y^2)^2 + (4xy)^2 + 8xy(x^2+y^2)
then get root( u)*u2

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alternative :

follow pascal 3_angle for fifth order ,i.e,

u2 = (a+b)^2 = a^2 +2ab+b^2
co-ef. = [1,2,1]

u3 = (a+b)^3 = a^3+3a^2b+3ab^2+b^3
co-ef. = [1,3,3,1]

u5 = (a+b)^5 ... co-ef. = [1,5,10,10,5,1]
in your case a = x , b = 2y

Answer 8

5C2 * (x)³ * (2y)²
10 * x³ * 4 * y²
= 40x³y²

Answer 9

I would use Pascal's triangle and get...
40x³y²