Any simpler method to expand (a+b+c+d)^4 ?

I find it extremely complicated when expand this ,please help!

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  • 1 decade ago
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    In some sense, there is no nice way of writing down this product. It has 35 terms; even if you had a very clever way to compute them, still have to write them all down!

    What can be done rather easily is to say what any particular term in the product must be. Let me explain what I mean. One way of thinking about the product

    (a+b+c+d)(a+b+c+d)(a+b+c+d)(a+b+c+d)

    is as follows: we consider every way of choosing one letter from the first parenthesized term, one from the second, one from the third, and one from the fourth. Once we've made those choices, we multiply them together, and add them all up.

    Consider the simpler case of the product (a+b)(a+b). We can choose to multiply a in the first term by a in the second, a in the first term by b in the second, b in the first term by a in the second, and b in the first term by b in the second. The result is

    aa + ab + ba + bb.

    Of course, this expansion isn't quite what we want; we haven't used the fact that ab and ba are the same, so in fact we can write it as a^2 + 2ab + b^2. The simplification in this case was easy since there were only two terms that could be added together.

    Your product has a similar big expansion, only much longer:

    aaaa+aaab+aaac+aaad+aaba+aabb+aabc+aabd+aaca+aacb+...

    If you think about it for a while, you could certainly list all the elements in this expansion in a routine way--- but it's a long list, with 256 terms. And we then have the considerable problem of simplifying it. For some terms it's easy: aaaa only appears once, for example, so we know the product is

    (a+b+c+d)^4 = a^4 + [terms that do not involve a^4].

    For other terms it's harder: aaab, for example, is the same as aaba and abaa and baaa, but these are the only terms of this type that would be in the above sum, so in fact

    (a+b+c+d)^4 = 4 a^3 b + [terms that do not involve a^3 b].

    Other terms are harder still: for example aabc is equal to aacb and abca and abac and bcaa and... as you can see, keeping track of all the different ways this one form will appear in the product (and hence figuring out what the a^2 bc term in (a+b+c+d)^4 is) becomes very difficult.

    What _can_ we say in general? Well, the product, when it _is_ simplified, will only have terms of the form

    a^( ) b^( ) c^( ) d^( )

    with the ( )'s replaced with certain powers (some of them can perhaps be zero, for instance the a^4 term has this form a^4 b^0 c^0 d^0).

    In fact, we can say more: the powers have to add up to 4. So there will be an a^4 term, and an a^3 b term, and an a^2 bc term, but no a^2bcd, and no c^5 term, or anything like that.

    There is a formula that tells us to figure out what the number in front of a^i b^j c^k d^l will be. It is 4! divided by i! j! k! l!.

    Here ! denotes the factorial function: 4! is 4*3*2*1 = 24, 3! is 3*2*1 = 6, 2! = 2*1 = 2, and 1! = 1, and 0! = 1 by definition.

    Example: the number in front of a^4 = a^4 b^0 c^0 d^0 is 4! divided by 4! 0! 0! 0!, which is 24 divided by 24*1*1*1, which is 1.

    The number in front of a^3 b = a^3 b^1 c^0 d^0 is

    4! divided by 3! 1! 0! 0!, or 24 divided by 6*1*1, or 4.

    The number in front of a^2 bc = a^2 b^1 c^1 d^0 is

    4! divided by 2! 1! 1! 0!, or 24 divided by 2*1*1*1, or 12.

    The number in front of abcd = a^1 b^1 c^1 d^1 is

    4! divided by 1! 1! 1! 1! or 24 divided by 1 or 24.

    And so on.

    This formula will tell us, given any particular term we would like to know about in the product, what number will be in front of it. But there are 35 terms to consider: a^4, a^3 b, a^3 c, a^3 d, a^2 b^2, a^2 c^2, a^2 d^2, a^2 bc, a^2 bd, a^2 cd, a b^3, ac^3, ad^3, ab^2 c, ab^2 d, on and on. This nice formula gives us the numbers in front of these terms, but there are so many of them that the expansion is still a pain to write by hand.

    For more information on the formula I gave above and why it works, see the wikipedia article for multinomial coefficients: http://en.wikipedia.org/wiki/Multinomial_theorem

  • Anonymous
    4 years ago

    Congrats on being a studious scholar in the pursuit of hedonism. Many are called - few are chosen!

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