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anonymous

  • one year ago

Can anyone please explain The Binomial Theorem and Pascal’s Triangle to me? I will medal?!

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  1. anonymous
    • one year ago
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    Practice problems would be great too because I have no idea how to do it

  2. ganeshie8
    • one year ago
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    First of all, recall that a "binomial" is a polynomial with "two" terms. so the theorem must have something to do with "two terms".

  3. anonymous
    • one year ago
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    Ok:)

  4. ganeshie8
    • one year ago
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    Indeed it does, the binomial theorem gives a nice expression for repeated multiplication of a binomial with itself: \[(x+y)(x+y)(x+y)\cdots\]

  5. anonymous
    • one year ago
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    I see, I see

  6. ganeshie8
    • one year ago
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    Without beating around the bush, here is the theorem in all its glory : \[\large (x+y)^n = \sum\limits_{k=0}^n \color{red}{\binom{n}{k}}x^ky^{n-k}\]

  7. anonymous
    • one year ago
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    I'm just gonna go ahead and do pascals triangle. In very simple words, its a number triangle. To build the triangle, start with "1" at the top, then continue placing numbers below it in a triangular pattern. Each number is the two numbers above it added together (except for the edges, which are all "1").

  8. anonymous
    • one year ago
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    :O Stop! The formula was NOT what I was expecting...

  9. anonymous
    • one year ago
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    But how does it relate to the binomial theorem

  10. ganeshie8
    • one year ago
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    \(\large \color{red}{\binom{n}{k}}\) is called "binomial coefficient", pascal's triangle helps you in finding this coefficeint.

  11. anonymous
    • one year ago
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    Let’s explore the first couple of powers of the binomial (a + b). (a + b)0 = 1 *Remember, anything to the zero power equals 1. (a + b)1 = a + b (a + b)2 = (a + b)(a + b) = a^2 + 2ab + b^2 (a + b)3 = (a + b)(a2 + 2ab + b2) = a^3 + 3a^2b + 3ab^2 + b^3

  12. anonymous
    • one year ago
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    uh-huh I'm processing this slowly, but I feel like I'm getting it

  13. anonymous
    • one year ago
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    Now take a look at the pascals triangle. (just search it up online)

  14. anonymous
    • one year ago
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    compare the coefficients, and powers with the pascals triangle

  15. anonymous
    • one year ago
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    1 a + b a^2 + 2ab + b^2 a^3 + 3a^2b + 3ab^2 + b^3

  16. anonymous
    • one year ago
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    and 1 11 121 1331

  17. anonymous
    • one year ago
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    Yes, I understand that.

  18. anonymous
    • one year ago
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    Can any of you guys give me like a practice question to help maybe?

  19. anonymous
    • one year ago
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    The row represents the coefficients

  20. anonymous
    • one year ago
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    Hold on ffggff i just need to say one more thing

  21. anonymous
    • one year ago
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    kk:)

  22. anonymous
    • one year ago
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    Take a look at the exponents for (a+b)^3 The powers of a decrease from 3 to 0, and the powers of b increase from 0 to 3.

  23. anonymous
    • one year ago
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    so that is the relationship between the two Here is a practice problem Expand (x + 2)^5 using the Binomial Theorem and Pascal’s triangle.

  24. anonymous
    • one year ago
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    Side note, is that the fifth row of pascals triangle would represent (a+b)^4, and the third would be (a+b) ^2. The reason is that the first row, represents (a+b)^0

  25. anonymous
    • one year ago
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    Ok, I'm lost lol. So I drew the little triangle thingy and got 1, 5, 10, 10, 5, 1. But now what?

  26. anonymous
    • one year ago
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    Like I can do the triangle but I don't know what to do afterwards?

  27. anonymous
    • one year ago
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    The sixth row of Pascal’s triangle would be 1, 5, 10, 10, 5, 1 and the exponents would start and end with 5. And that is what you got

  28. anonymous
    • one year ago
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    Yup

  29. anonymous
    • one year ago
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    So then \[1x^5 2^0 + 5x^4 2^1 + 10x^3 2^2 + 10x^2 2^3 + 5x^12^4 + 1x^02^5 =\] \[x^5 + 5 • 2x^4 + 10 • 4x^3 + 10 • 8x^2 + 5 • 16x + 32 =\] \[x^5 + 10x^4 + 40x^3 + 80x^2 + 80x + 32\]

  30. anonymous
    • one year ago
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    :O Whoa! Okay! How did you add the 6th row of the triangle to the equation?

  31. anonymous
    • one year ago
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    I mean, how did you get those exponents?

  32. anonymous
    • one year ago
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    Remember this? Take a look at the exponents for (a+b)^3 The powers of a decrease from 3 to 0, and the powers of b increase from 0 to 3. SO basically, as it goes, the powers of x decrease from the n to 0 (where n is the exponent, or (x+y)^n) And the powers of the other one (in this case it is 2) increases from 0 to n.

  33. anonymous
    • one year ago
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    Ok :) Thank-you, I get it now!!!!

  34. anonymous
    • one year ago
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    NP!!! :)

  35. anonymous
    • one year ago
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    So basically they are coming from (a+b)^5 in a way?

  36. anonymous
    • one year ago
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    @zzr0ck3r I feel so bad! Lol you have been typing forever :)

  37. anonymous
    • one year ago
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    well, yes, So to finalize it let the equation be (a+b)^n So a's exponents would go from n to 0 and b's would increase from 0 to n.

  38. anonymous
    • one year ago
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    how do you what to do to break it up into 5? Like how the first one is 1x^5 2^0 and the second one is 5x^42^1?

  39. zzr0ck3r
    • one year ago
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    A neat trick you can use instead of the triangle. Example: \((a+b)^4\) We know that if will look something like this \(\text{_}a^4+\text{_}a^3b+\text{_}a^2b^2+\text{_}ab^3+\text{_}b^4\) Now the question is, what do we put in the _ places Well we know the first coefficient will be \(1\). For the second _ we multiply the power of \(a\) in the first term, by the coefficient in the first term and then divide that all by \(1\) (we use one because the previous two steps were in the first term). So \((4*1)/1=4\) and that is our second coefficient. so we have \(a^4+4a^3b+\text{_}a^2b^2+\text{_}ab^3+\text{_}b^4\) For the third coefficient, we multiply the power of \(a\) in the second term by the coefficient in the second term and divide by \(2\) (we are doing stuff in the second term now) and we get \(3*4/2 = 6\) and that is our third coefficient. Similarly our fourth coefficient is \(2*6/3=4\) and our last coefficient is \(1*4/4=1\) So we have \(a^4+4a^3b+6a^2b^2+4ab^3+b^4\). Also I am sure someone pointed this out above, but once we find the first half of the coefficients, we are done (notice the pattern). Ok this all seems very hard and long, but that was just because I was doing lots of steps. If wanted to do \((a+b)^7\) First we quickly write out \(a^7+a^6b+a^5b^2+a^4b^3+a^3b^4+a^2b^5+ab^6+b^7\) and note the leading coefficient is \(1\). then we do \(7*1/1=7\\ 6*7/2=21\\ 5*21/3=5*7=35\) and we are done because we have gone half way so we have \((a+b)^7=a^7+7a^6b+21a^5b^2+35a^4b^3+35a^3b^4+21a^2b^5+7ab^6+b^7\)

  40. anonymous
    • one year ago
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    do you see what I mean kinda?

  41. zzr0ck3r
    • one year ago
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    I did not realize it was going to be so long once I started...

  42. anonymous
    • one year ago
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    @zzr0ck3r Thank-you :), I might just use your way from now on. It's simple yet very effective!

  43. anonymous
    • one year ago
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    well yes @zzr0ck3r but ffggfgf wants to know the relationship between binomial theorem and pascals triangle.

  44. zzr0ck3r
    • one year ago
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    this actually shows it:)

  45. anonymous
    • one year ago
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    It's a good method.

  46. zzr0ck3r
    • one year ago
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    I can generally do these things and write them down faster than someone can put it in a calculator and write it down.

  47. anonymous
    • one year ago
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    i use a different method than these, it is quite fast also.

  48. zzr0ck3r
    • one year ago
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    But anyway, it was meant to be a little extra add-on to what you all were saying...

  49. anonymous
    • one year ago
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    i would write it, but i don't want to spend another hour lol

  50. anonymous
    • one year ago
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    Thanks to everyone who helped!! I really understand how to do it now!! :)

  51. ganeshie8
    • one year ago
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    I think that works because of combinations, which term we pick in each binomial : |dw:1436843111371:dw|

  52. misty1212
    • one year ago
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    one day see if you can find a nice proof of why pascal's triangle works for these questions, and also why the construction of the triangle give the entries \(\binom{n}{k}\) someday later on i mean

  53. anonymous
    • one year ago
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    Thanks misty1212 :) Lol once I get to Ap cal I will!

  54. anonymous
    • one year ago
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    Then what do i do ganeshie?

  55. ganeshie8
    • one year ago
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    You will know what to do once you're faced with an example problem. Let me see if I can cookup an interesting problem :)

  56. anonymous
    • one year ago
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    Kinda scared now hahah :) But I will try

  57. ganeshie8
    • one year ago
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    Find the coefficient of \(x^6\) when you expand out below product : \[(x+1)(x+1)(x+1)(x+1)(x+1)(x+1)(x+1)\] (there are 7 terms in the product)

  58. anonymous
    • one year ago
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    Ok, hold on please

  59. anonymous
    • one year ago
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    Would it be 1?

  60. ganeshie8
    • one year ago
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    nope, but how did you guess 1 ?

  61. anonymous
    • one year ago
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    1 9 36 84 126 126 84 36 9 1

  62. anonymous
    • one year ago
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    I did the (a+b)^6 using the triangle

  63. anonymous
    • one year ago
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    the first one would be one, as the first one would be x^7.

  64. anonymous
    • one year ago
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    remember what i said the exponents on the x part decrease from n. And btw, @ganeshie8 typed the expanded version of (x+1)^7

  65. anonymous
    • one year ago
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    ohhh whoops But yes, go on.

  66. ganeshie8
    • one year ago
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    Yes, maybe forget about the theorem/pascal's triangle and try working it pretending we never heard of these

  67. anonymous
    • one year ago
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    tru tru, lets forget lol

  68. dan815
    • one year ago
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    @misty1212 you wanted a proof for why the construction of pascal triangles has combinatoric solutions?

  69. ganeshie8
    • one year ago
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    \[(x+1)(x+1)(x+1)(x+1)(x+1)(x+1)(x+1)\] Notice that there are \(7\) factors here and since we want the coefficient of \(x^6\), we must pick "x" from "6" factors and pick "1" from one factor. We can pick "1" from any of the 7 factors in exactly 7 ways. so the coefficeint of \(x^6\) in above product is simply \(7\)

  70. anonymous
    • one year ago
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    Ok!

  71. ganeshie8
    • one year ago
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    As you can see, the "coefficient" is really equivalent to "number of ways of choosing a factor"

  72. anonymous
    • one year ago
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    so would it be 1 7 21 35 35 21 7 1

  73. anonymous
    • one year ago
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    Yup!

  74. ganeshie8
    • one year ago
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    If you use pascal's triangle, yes.

  75. anonymous
    • one year ago
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    Gotcha

  76. ganeshie8
    • one year ago
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    Let me ask you a side question : Suppose you went for shopping and liked 7 different shirts, but you can only buy 6 shirts. how many total ways are there to choose 6 shirts from 7 shirts ?

  77. dan815
    • one year ago
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    |dw:1436844308425:dw|

  78. anonymous
    • one year ago
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    6

  79. ganeshie8
    • one year ago
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    think again

  80. anonymous
    • one year ago
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    6x7 basically

  81. ganeshie8
    • one year ago
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    Let me change the question slightly : There are 7 shirts and you can buy only 1 shirt. How many ways are there to choose 1 shirt from 7 shirts ?

  82. anonymous
    • one year ago
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    7? I don't really understand the question. Can u just explain it to me please?

  83. ganeshie8
    • one year ago
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    Correct! I will explain after the drill in the end. lets try another related question : There are 7 shirts and you can buy only 1 shirt. How many ways are there to "reject" 6 shirts from 7 shirts ?

  84. anonymous
    • one year ago
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    7?

  85. ganeshie8
    • one year ago
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    Yes you do see that "choosing 1 shirt from 7 shirts" is same as "rejecting 6 shirts from 7 shirts"

  86. anonymous
    • one year ago
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    yup. I do see.

  87. anonymous
    • one year ago
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    At first I didn't, but now I do.

  88. ganeshie8
    • one year ago
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    "choosing 1 girlfriend from available 7 friends" is same as "rejecting 6 girls from the available 7 friends"

  89. ganeshie8
    • one year ago
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    sry for bad analogy but you get the point

  90. anonymous
    • one year ago
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    Lol yeah, I get it :D

  91. ganeshie8
    • one year ago
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    so what has that anything to do with finding the coefficient of \(x^6\) in the below product : \[(x+1)(x+1)(x+1)(x+1)(x+1)(x+1)(x+1)\] ?

  92. anonymous
    • one year ago
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    oh, okay so I'm assuming it would be 7

  93. anonymous
    • one year ago
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    Yes! Omg! Ok, I see. Visuals really do help!!

  94. ganeshie8
    • one year ago
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    Maybe writing the product pictorially helps in seeing whats going on : |dw:1436845412826:dw|

  95. ganeshie8
    • one year ago
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    I feel that is enough for one session, just open a new thread if you still have any question :) good luck!

  96. anonymous
    • one year ago
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    Thanks for all your help<3

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