Question

Use Pascal's Triangle to expand the binomial.
(d – 3)6

Answer choices:
a)d6 – 18d5 + 135d4 – 540d3 + 1,215d2 – 1,458d + 729

b)d6 + 18d5 + 135d4 + 540d3 + 1,215d2 + 1,458d + 729

c)d6 – 6d5 + 15d4 – 20d3 + 15d2 – 6d + 1

d)d6 + 6d5 + 15d4 + 20d3 + 15d2 + 6d + 1

Answer 1

do you know what the sixth layer of pascal's triangle looks like?
http://www.mathsisfun.com/pascals-triangle.html

Answer 2

ok? I don't understand how to do this like at all..

Answer 3

go down to the sixth level

1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1

Answer 4

is it clear how the triangle is constructed?

Answer 5

yeah.

Answer 6

ok so we read off the coefficients
1 6 15 20 15 6 1

Answer 7

each term will look like
\[(d – 3)^6 =1\times b^6+5\times b^5\times (-3)+15\times b^2(-3)^4\]\[+20\times b^3(-3)^3+15\times b^2(-3)^4+6\times b^1(-3)^5+(-3)^6\]

Answer 8

damn typo

Answer 9

where do the letters come from though? cause their is obviously no letters in the triangle..

Answer 10

if you have \(a+b)^n\) each term looks like \(c_k(a^k)(b^{n-k})\) the exponents have to add up to \(n\)and you find the number out front (the coefficient) from pascal's triangle

Answer 11

i thought all i had to do is add the two top numbers to get the bottom number..?

Answer 12

6C0(d)^6(-3)^0 + 6C1(d)^5(-3) + 6C2(d)^4(-3)^2 + 6C3(d)^3(-3)^3 + 6C4(d)^2(-3)^4 + 6C5(d)(-3)^5 + 6C6(d)^0(-3)^6
= ...... good luck