# Expanding

• Aug 21st 2009, 10:54 AM
rel85
Expanding
Hello,

1) (a+b)^6
2) (1-x)^8

Regards,
• Aug 21st 2009, 11:00 AM
stapel
Apply the Binomial Theorem. (Wink)
• Aug 21st 2009, 11:13 AM
Chris L T521
Quote:

Originally Posted by rel85
Hello,

1) (a+b)^6
2) (1-x)^8

Regards,

Recall Pascal's Triangle:

Code:

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
1  8  28  56  70  56  28  8  1

So when you go to expand $\displaystyle (a+b)^6$, the coefficients of your terms will be from the 7th row of Pascal's Triangle.

Code:

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
1  8  28  56  70  56  28  8  1

Thus, it follows that $\displaystyle (a+b)^6=1\cdot a^6b^0+6\cdot a^5b^1+15\cdot a^4b^2+20\cdot a^3b^3+15\cdot a^2b^4+6\cdot a^1b^5+1\cdot a^0b^6$

A similar process is done for $\displaystyle \left(1-x\right)^8=\left(1+\left(-x\right)\right)^8$. You just look at the 9th row of the Pascal's Triangle for the coefficients.

Can you try the last one?
• Aug 21st 2009, 11:50 AM
rel85
Quote:

Originally Posted by Chris L T521
Recall Pascal's Triangle:

Code:

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
1  8  28  56  70  56  28  8  1

So when you go to expand $\displaystyle (a+b)^6$, the coefficients of your terms will be from the 7th row of Pascal's Triangle.

Code:

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
1  8  28  56  70  56  28  8  1

Thus, it follows that $\displaystyle (a+b)^6=1\cdot a^6b^0+6\cdot a^5b^1+15\cdot a^4b^2+20\cdot a^3b^3+15\cdot a^2b^4+6\cdot a^1b^5+1\cdot a^0b^6$

A similar process is done for $\displaystyle \left(1-x\right)^8=\left(1+\left(-x\right)\right)^8$. You just look at the 9th row of the Pascal's Triangle for the coefficients.

Can you try the last one?

Okay I see it. How do I go about using that neat notation? Otherwise the answer is going to be extremely messy!

Can the first answer be simplified, or would the answer you gave, be the final step? Is there any other 'types' of binomial question that requires a different method?

Regards,
• Aug 21st 2009, 12:01 PM
Chris L T521
Quote:

Originally Posted by rel85
Okay I see it. How do I go about using that neat notation? Otherwise the answer is going to be extremely messy!

Can the first answer be simplified, or would the answer you gave, be the final step? Is there any other 'types' of binomial question that requires a different method?

Regards,

You could do one more step to get $\displaystyle \left(a+b\right)^6=a^6+6a^5b+15a^4b^2+20a^3b^3+15a ^2b^4+6ab^5+b^6$

With regards to neat notation, do you mean the math typesetting? If so, we use $\displaystyle \text{\LaTeX}$ here on the forums. See the LaTeX Help subforum (in particular, the two LaTeX tutorials) to learn the basics.

With regards to other binomial type of questions, there are some variations that ask you to exand things like $\displaystyle \left(a+b\right)^{r};\,\, r\in\mathbb{C}$ (r doesn't have to be an integer power; it can be complex [or real]). You can read more about it here: Binomial theorem - Wikipedia, the free encyclopedia
• Aug 21st 2009, 12:54 PM
rel85
Quote:

Originally Posted by Chris L T521
You could do one more step to get $\displaystyle \left(a+b\right)^6=a^6+6a^5b+15a^4b^2+20a^3b^3+15a ^2b^4+6ab^5+b^6$

With regards to neat notation, do you mean the math typesetting? If so, we use $\displaystyle \text{\LaTeX}$ here on the forums. See the LaTeX Help subforum (in particular, the two LaTeX tutorials) to learn the basics.

With regards to other binomial type of questions, there are some variations that ask you to exand things like $\displaystyle \left(a+b\right)^{r};\,\, r\in\mathbb{C}$ (r doesn't have to be an integer power; it can be complex [or real]). You can read more about it here: Binomial theorem - Wikipedia, the free encyclopedia

Thank you so much, specifically I was wondering if there was any subtract questions... or can they all be manipulated from $\displaystyle (a - b)^x = (a + (-b)^x$

With regards to the second attempt:

$\displaystyle (1 + (-x))^8$

9th line = 1, 9, 34, 84, 126, 126, 84, 34, 9, 1

so: $\displaystyle 1 . 1^8(-x)^0 + 1 . 9^7(-x)^1 + 1 . 34^6(-x)^2 + 1 . 84^5(-x)^3 + 1 . 126^4(-x)^4$
$\displaystyle + 1 . 126^3(-x)^5 + 1 . 84^2(-x)^6 + 1 . 34^1(-x)^7 + 1 . 9^0(-x)^8 + 1$

Is this okay?
• Aug 21st 2009, 02:00 PM
Chris L T521
Quote:

Originally Posted by Chris L T521
Recall Pascal's Triangle:

Code:

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
1  8  28  56  70  56  28  8  1

So when you go to expand $\displaystyle (a+b)^6$, the coefficients of your terms will be from the 7th row of Pascal's Triangle.

Code:

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
1  8  28  56  70  56  28  8  1

Thus, it follows that $\displaystyle (a+b)^6=1\cdot a^6b^0+6\cdot a^5b^1+15\cdot a^4b^2+20\cdot a^3b^3+15\cdot a^2b^4+6\cdot a^1b^5+1\cdot a^0b^6$

A similar process is done for $\displaystyle \left(1-x\right)^8=\left(1+\left(-x\right)\right)^8$. You just look at the 9th row of the Pascal's Triangle for the coefficients.

Can you try the last one?

Quote:

Originally Posted by rel85
Thank you so much, specifically I was wondering if there was any subtract questions... or can they all be manipulated from $\displaystyle (a - b)^x = (a + (-b)^x$

With regards to the second attempt:

$\displaystyle (1 + (-x))^8$

9th line = 1, 9, 34, 84, 126, 126, 84, 34, 9, 1

so: $\displaystyle 1 . 1^8(-x)^0 + 1 . 9^7(-x)^1 + 1 . 34^6(-x)^2 + 1 . 84^5(-x)^3 + 1 . 126^4(-x)^4$
$\displaystyle + 1 . 126^3(-x)^5 + 1 . 84^2(-x)^6 + 1 . 34^1(-x)^7 + 1 . 9^0(-x)^8 + 1$

Is this okay?

You used the coefficients from the tenth row!

"1" is the first row, "1 1" is the second, "1 2 1" is the third, and so on.

The ninth line is "1 8 28 56 70 56 28 8 1"

So it follows that $\displaystyle \left(1+\left(-x\right)\right)^{8}=1\cdot1^8\left(-x\right)^0+8\cdot1^7\left(-x\right)^1+28\cdot1^6\left(-x\right)^2+56\cdot1^5\left(-x\right)^3+70\cdot1^4\left(-x\right)^4$ $\displaystyle +56\cdot1^3\left(-x\right)^5+28\cdot1^2\left(-x\right)^6+8\cdot1^1\left(-x\right)^7+1\cdot1^0\left(-x\right)^8$

That simplifies to $\displaystyle 1-8x+28x^2-56x^3+70x^4-56x^5+28x^6-8x^7+x^8$.
• Aug 21st 2009, 02:15 PM
rel85
Quote:

Originally Posted by Chris L T521
You used the coefficients from the tenth row!

"1" is the first row, "1 1" is the second, "1 2 1" is the third, and so on.

The ninth line is "1 8 28 56 70 56 28 8 1"

So it follows that $\displaystyle \left(1+\left(-x\right)\right)^{8}=1\cdot1^8\left(-x\right)^0+8\cdot1^7\left(-x\right)^1+28\cdot1^6\left(-x\right)^2+56\cdot1^5\left(-x\right)^3+70\cdot1^4\left(-x\right)^4$ $\displaystyle +56\cdot1^3\left(-x\right)^5+28\cdot1^2\left(-x\right)^6+8\cdot1^1\left(-x\right)^7+1\cdot1^0\left(-x\right)^8$

That simplifies to $\displaystyle 1-8x+28x^2-56x^3+70x^4-56x^5+28x^6-8x^7+x^8$.

Careless of me!