Question about HP 10bII Calculator
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Using sigma notation, and factorials for the combinatorial numbers, here is the binomial theorem:
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.
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.
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.
This is the first term or, first coefficient 6*x^5*y, when k=1.
Solution so far = x^6+6*x^5*y
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
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
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
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
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
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
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