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How to solve binomial theorem with ti 84 plus calculator

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oneplusgh
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SOURCE: TI-89 Titanium: I want to

Using elementary algebria in the binomial theorem, I expanded the power (x + y)^n into a sum involving terms in the form a x^b y^c. The coefficient of each term is a positive integer, and the sum of the exponents of x and y in each term is n. This is known as binomial coefficients and are none other than combinatorial numbers.

Combinatorial interpretation:

Using binomial coefficient (n over k) allowed me to choose k elements from an n-element set. This you will see in my calculations on my Ti 89. This also allowed me to use (x+y)^n to rewrite as a product. Then I was able to combine like terms to solve for the solution as shown below.
(x+y)^6= (x+y)(x+y)(x+y)(x+y)(x+y)(x+y) = x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6

This also follows Newton's generalized binomial theorem:


oneplusgh_15.jpg
Now to solve using the Ti 89.


Using sigma notation, and factorials for the combinatorial numbers, here is the binomial theorem:

oneplusgh.gif

The summation sign is the general term. Each term in the sum will look like that as you will see on my calculator display. Tthe first term having k = 0; then k = 1, k = 2, and so on, up to k = n.
Notice that the sum of the exponents (n ? k) + k, always equals n.



oneplusgh_26.jpg


The summation being preformed on the Ti 89. The actual summation was preformed earlier. I just wanted to show the symbolic value of (n) in both calculations. All I need to do is drop the summation sign to the actual calculation and, fill in the term value (k), for each binomial coefficient.



oneplusgh_18.jpg

This is the zero th term. x^6, when k=0. Notice how easy the calculations will be. All I'm doing is adding 1 to the value of k.


oneplusgh_19.jpg

This is the first term or, first coefficient 6*x^5*y, when k=1.
Solution so far = x^6+6*x^5*y



oneplusgh_20.jpg


This is the 2nd term or, 2nd coefficient 15*x^4*y^2, when k=2.
Solution so far = x^6+6*x^5*y+15*x^4*y^2



oneplusgh_27.jpg



This is the 3rd term or, 3rd coefficient 20*x^3*y^3, when k=3.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3



oneplusgh_28.jpg



This is the 4th term or, 4th coefficient 15*x^2*y^4, when k=4.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4



oneplusgh_21.jpg


This is the 5th term or, 5th coefficient 6*x*y^5, when k=5.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5



oneplusgh_22.jpg

This is the 6th term or, 6th coefficient y^6, when k=6.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5+y^6



oneplusgh_23.jpg

Putting the coefficients together was equal or, the same as for when I used the expand command on the Ti 89.

binomial coefficient (n over k) for (x+y)^6
x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5+y^6












Posted on Jan 04, 2011

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1 Answer

Where is the '!' button on the TI 84 plus calculator?


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(3x-2y)^4 use the binomial theorem


You have to type expand((38x-28y)^4,x). See captured images
2_9_2012_10_43_20_am.jpg

Jan 04, 2012 | Texas Instruments TI-89 Calculator

2 Answers

TI-89 Titanium: I want to solve a Binomial Theorem problem (x+y)^6 how would i go about solving this in the calculator?


Using elementary algebria in the binomial theorem, I expanded the power (x + y)^n into a sum involving terms in the form a x^b y^c. The coefficient of each term is a positive integer, and the sum of the exponents of x and y in each term is n. This is known as binomial coefficients and are none other than combinatorial numbers.

Combinatorial interpretation:

Using binomial coefficient (n over k) allowed me to choose k elements from an n-element set. This you will see in my calculations on my Ti 89. This also allowed me to use (x+y)^n to rewrite as a product. Then I was able to combine like terms to solve for the solution as shown below.
(x+y)^6= (x+y)(x+y)(x+y)(x+y)(x+y)(x+y) = x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6

This also follows Newton's generalized binomial theorem:


oneplusgh_15.jpg
Now to solve using the Ti 89.


Using sigma notation, and factorials for the combinatorial numbers, here is the binomial theorem:

oneplusgh.gif

The summation sign is the general term. Each term in the sum will look like that as you will see on my calculator display. Tthe first term having k = 0; then k = 1, k = 2, and so on, up to k = n.
Notice that the sum of the exponents (n ? k) + k, always equals n.



oneplusgh_26.jpg


The summation being preformed on the Ti 89. The actual summation was preformed earlier. I just wanted to show the symbolic value of (n) in both calculations. All I need to do is drop the summation sign to the actual calculation and, fill in the term value (k), for each binomial coefficient.



oneplusgh_18.jpg

This is the zero th term. x^6, when k=0. Notice how easy the calculations will be. All I'm doing is adding 1 to the value of k.


oneplusgh_19.jpg

This is the first term or, first coefficient 6*x^5*y, when k=1.
Solution so far = x^6+6*x^5*y



oneplusgh_20.jpg


This is the 2nd term or, 2nd coefficient 15*x^4*y^2, when k=2.
Solution so far = x^6+6*x^5*y+15*x^4*y^2



oneplusgh_27.jpg



This is the 3rd term or, 3rd coefficient 20*x^3*y^3, when k=3.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3



oneplusgh_28.jpg



This is the 4th term or, 4th coefficient 15*x^2*y^4, when k=4.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4



oneplusgh_21.jpg


This is the 5th term or, 5th coefficient 6*x*y^5, when k=5.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5



oneplusgh_22.jpg

This is the 6th term or, 6th coefficient y^6, when k=6.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5+y^6



oneplusgh_23.jpg

Putting the coefficients together was equal or, the same as for when I used the expand command on the Ti 89.

binomial coefficient (n over k) for (x+y)^6
x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5+y^6












Jan 04, 2011 | Texas Instruments TI-89 Calculator

1 Answer

{Use the binomial theorem to find the eight term of (3x-2y)^13}


This may help:
http://en.wikipedia.org/wiki/Binomial_theorem
Rate me, thanks.

Jun 19, 2009 | Texas Instruments TI-89 Calculator

1 Answer

Error message with binopdf function


Yes, eleven million is rather extreme for the binomial distribution. For this large a value the binomial distribution is sufficiently indistinguishable from the normal approximation.

Apr 15, 2009 | Texas Instruments TI-84 Plus Calculator

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