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# use pascals triangle to expand the binomial (d=3)^7?

i need help with this problem. im stuck!!!!

### 6 Answers

- Scarlet ManukaLv 71 decade agoFavorite Answer
Obviously, the first thing you need is seven rows (or, at least, the seventh row) of the triangle. If my mental calculations are correct (and it is midnight here, incidentally, so be warned!) the values you get should be these:

1 7 21 35 35 21 7 1

So the answer is

(d-3)^7

= 1.d^7 + 7.d^6.(-3) + 21.d^5.(-3)^2 + 35.d^4.(-3)^3 + 35.d^3.(-3)^4 + 21.d^2.(-3)^5 + 7.d.(-3)^6 + (-3)^7

= d^7 - 21d^6 + 189d^5 - 945d^4 + 2835d^3 - 5103d^2 + 5103d - 2187.

- 1 decade ago
I'm not sure if I understand the question, you talk about a binomial that usually has the form (a+b)^c, so I assume that there was a mispelling error and you tried to write: (d+3)^7.

If this is the case, then here is my answer: there is a property of the binomial expansion that says that when you want to expand to the Nth exponent (in your case, N = 7), you use the Nth level of the triangle of Pascal.

First of all, what is Pascal's triangle? Very easy. it is a triangle built as follows: you put a 1 at the top, and then you follow this rule: under each number, put one number at right and one number at left; the value of each number is given by the sum of its two "parents" in the pyramid.

Let's see it step by step:

1) we put a 1 at the top:

1

2) then we put two childs for this one; the value of each child is given by the value of its "parents", in this case the 1:

1

1 1

3) after this, we add a second line of "childs"; notice that the central child has the two parents "1" and "1", so its value will be 1+1 = 2

1

1 1

1 2 1

4) after this, we add a 3rd line of "child"; the 2 central childs having as "parents", 2 and 1, their value will be 3 = 2+1

1

1 1

1 2 1

1 3 3 1

5) then we add a 4th line of childs, notice that some of these childs will be under 1 and 3, so their value will be 4 = 1+3; and the central child is under 3 and 3, so its value will be 6=3+3

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

do you see the idea? we can continue adding lines to Pascal's triangle, until reaching the 7th level that is the one we'll need to use:

1

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

1 7 21 35 35 21 7 1

Here we are at 7th level: the external 7 are under 1 and 6 (7 = 1+6), the 21 is under 15+6, the 35 are under 15+20.

Now, a well known rule says that to expand a binomial with exponent N you just combine all the monomiums (formed by the multiplication of first term and second term with exponents such that its sum is N), and apply to each monomium the coefficients coming from pascal's triangle line given by the exponent.

What this means in easy words? it means that for example to expand:

(a+b)^3, you will combine these terms:

a^3, (a^2)*b, a*(b^2), b^3 (e.g. all the combinations of exponents of "a" and "b" that sum 3), and you will use the following coefficients for these terms: 1 3 3 1 (this is the 3rd line of Pascal's triangle), so you will have:

1*a^3

3*(a^2)*b

3*a*(b^2)

1*b^3

(see how 1 3 3 1 is combined with the 4 terms mentionned before)

and so:

(a+b)^3 = a^3 + 3*(a^2)*b + 3*a*(b^2) + b^3

Now, let's apply this idea to the problem you have to solve:

(d+3)^7

we have to combine the following terms:

d^7

(d^6)*3

(d^5)*(3^2)

(d^4)*(3^3)

(d^3)*(3^4)

(d^2)*(3^5)

d*(3^6)

3^7

(e.g. all the combinations of "d" and "3" with exponents that sum 7)

and we have to combine them with the coefficients coming from the 7th line of Pascal's triangle, e.g. 1 7 21 35 35 21 7 1

So, we obtain these terms:

1*(d^7)

7*(d^6)*3

21*(d^5)*(3^2)

35*(d^4)*(3^3)

35*(d^3)*(3^4)

21*(d^2)*(3^5)

7*d*(3^6)

1*(3^7)

And the addition of all these terms is the expantion of the binomial (d+3)^7, e.g.:

(d+3)^7 = d^7 + 7*(d^6)*3 + 21*(d^5)*(3^2) + 35*(d^4)*(3^3) + 35*(d^3)*(3^4) + 21*(d^2)*(3^5) + 7*d*(3^6) + 3^7

We can simplify the expression by observing that 3^2 = 9, 3^3 = 27, 3^4 = 81, 3^5 = 243, 3^6 = 729, 3^7 = 2177, but this I leave as a small task for you :)

- RaymondLv 71 decade ago
A binomial has the form (a+b)

An unexpanded binomial has the form (a+b(^7

Pascals's triangle has the form shown at the first web site.

The numbers are the factors of each term in the expansion. The first line (the single number 1) is called line 0.

Each number in the triangle is the sum of the two numbers above it. For example, in line number 5 (1 5 10 10 5 1) a 10 is the sum of 4 and 6.

The expansion of (a+b)^5 is

1 a^5 + 5 a^4 b + 10 a^3b^2 + 10 a^2b^3 + 5 a b^4 + 1 b^5

To solve your problem:

Find line number 7 by filling in the sums (i.e., do line 6 first, then line 7)

Then expand by using the receipe, remembering that if b=-3 then b^4=+81, b^5=-243 and so on.

Source(s): (1) http://en.wikipedia.org/wiki/Pascal's_Triangle - lheureuxLv 44 years ago
Pascals triangle is amazingly elementary as quickly as you get the carry close of what occurring: 2^6 + 6(2)^5(sqrt(5)) + 15(2)^4(sqrt(5))^2 + 20(2)^3(sqrt(5))^3 + 15(2)^2(sqrt(5))^4 + 6(2)(sqrt(5))^5 + (sqrt(5))^6 next you may desire to multiply it out: =sixty 4+ 192(sqrt(5)) + 1200 + 800(sqrt(5)) + 1500 + 3 hundred(sqrt(5)) + a hundred twenty five =2889+1292(sqrt(5)) =5778 wish that helps, Ptr

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