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# binomial theorem

Is the number of terms in any binomial expansion always one more than the degree?

So, (x + y) ^ 10 - number of terms is 11.   And also 11 for  (x + y +2z) ^ 10 ??

And for (w + 2x + y -5z) ^ 12 - number of terms is 13.

Also, I know how to find the coefficient of any term where there are only 2 factors, like:

(2s -t) ^ 12

But how do I find the coefficient of the x^2 y^3 z^5 term for (x + y + z) ^ 10 ?
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HLRosenberger
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1 Solution

Commented:
Is the number of terms in any binomial expansion always one more than the degree?
YES

So, (x + y) ^ 10 - number of terms is 11.
YES

And also 11 for  (x + y +2z) ^ 10 ??
NO.  This has three terms so it is not a binomial expansion.
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Commented:
http://en.wikipedia.org/wiki/Trinomial_expansion

But how do I find the coefficient of the x^2 y^3 z^5 term for (x + y + z) ^ 10 ?
10!/(2! 3! 5!)  =  2520
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Commented:
From the Pythagorean Pyramid:

(x + y +2z) ^ 10    will have 10*(10+1)/2  =  55 terms

(w + 2x + y -5z) ^ 12 -  will have 12*(12+1)/2  =  78 terms
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Author Commented:
d-glitchP

How do I find the number of term using a formula?
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Commented:
The formula for trinomial coefficients is in the Wikipedia article.

I used it to find one of the 55 terms in the expansion of    (x + y + z) ^ 10

2520 * x² * y³ * z^5

I can find the coefficient for any other term you specify the same way.

What exactly are you trying to do?
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Commented:
Think about the Pythagorean Pyramid and the expansion of  (x + y + z) ^ n

The top (0th) level of the Pyramid has one term:         (x + y + z) ^ 0  =  1

The next (1st) level of the Pyramid has three terms:   (x + y + z) ^ 1  =  x + y + z

The next (2nd) level of the Pyramid has six terms:
(x + y + z) ^ 2  =  x² + 2xy + 2xz + y² + 2yz + z²
==========================================================================

Picture each level of the Pyramid as a triangle with x at the top, y at the bottom left, and z at the bottom right.

The 10th level of the Pyramid has 55 terms.  The triangle that makes up this level has 11 rows.

The frist row has one term:          x^10

The second row has two terms:   x^9 y  +  x^9 z

[Starting here I am ignoring coefficients.  Use the formula.]

The third row has three terms:    x^8 y²  +  x^8 y z  +  x^8 z²
.  .  .  .

The tenth row has ten terms:       x y^9  +  x y^8 z  +  . . .  +  x y z^8  +  x z^9

The last row has eleven terms:     y^10  +   y^9 z    +  . . .  +   y z^9   +    z^10
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Author Commented:
Thanks so Much!
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Author Commented:
One more question for :

(w + 2x + y -5z) ^ 12 -  will have 12*(12+1)/2  =  78 terms

How did you come up with 12*(12+1)/2?  What's the general formula the number of terms for a multinomial expansion?
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Commented:
(w + 2x + y -5z) ^ 12 -  will have 12*(12+1)/2  =  78 terms

No.    It will have.   78 +66 +55 +45 +36 +21 +15 +10 +6 +3 +1  terms.

(x)^n   will have 1  term

(x + y)^n   will have n+1  terms

(x + y +z)^n   will have (n+1)(n+2)/2  terms.
This is the sum of the integers from 1 to n+1.  These are triangonal numbers.

(w + x + y +z)^n   will have on the order of n^3 terms.
It will be the sum of the first n+1 triangonal numbers.  These are pyramidal numbers.

Each higher level will build on the one below.
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Commented:
Correction

No.    It will have.   78 +66 +55 +45 +36 +28 +21 +15 +10 +6 +3 +1  terms.
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Commented:
More corrections:

(x + y +2z) ^ 10    will have  (n+1)(n+2)   =   11*(11+1)/2  =  66 terms

(w + 2x + y -5z) ^ 12 -  will have 13*(13+1)/2  =  91 terms
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