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mathmath333

  • one year ago

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  1. abb0t
    • one year ago
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    answer

  2. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{Find the number of terms in }\hspace{.33em}\\~\\ & (a+b+c+d)^{50} \hspace{.33em}\\~\\ \end{align}}\)

  3. anonymous
    • one year ago
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    4 or 5?

  4. imqwerty
    • one year ago
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    have u tried using binomial :)

  5. mathmath333
    • one year ago
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    how ?

  6. imqwerty
    • one year ago
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    u can see this if u want- http://mathforum.org/library/drmath/view/68607.html :)

  7. mathmath333
    • one year ago
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    \(\LARGE { 53\choose 3}\)

  8. mathmath333
    • one year ago
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    this is correct?

  9. imqwerty
    • one year ago
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    yes :)

  10. mathmath333
    • one year ago
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    r u typing

  11. freckles
    • one year ago
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    I got mine answer a bit longer way... \[(a+b+c+d)^{50} \\ =(a+b)^{50}+c_1(a+b)^{49}(c+d)+c_2(a+b)^{48}(c+d)^2+ \cdots \\ +c_{24}(a+b)^{26}(c+d)^{24}+c_{25}(a+b)^{25}(c+d)^{25}+c_{26}(a+b)^{24}(c+d)^{26} + \cdots \\+ c_{48}(a+b)^{2}(c+d)^{48}+c_{49}(a+b)(c+d)^{49}+(c+d)^{50} \\ \\ \text{ so we have terms of that thingy } = \\ 51+50(2)+49(3)+\cdots+27(25)+26(26)+25(27)+\cdots +3(49)+2(50)+51\] \[=2 \cdot \sum_{i=1}^{25}i(52-i)+26 \cdot 26 \\ =2 \sum_{i=1}^{25}(52i-i^2)+26 \cdot 26 \\ =2 ( 52 \cdot \frac{25(25+1)}{2}- \frac{25(25+1)(2 \cdot 25+1)}{2})+26 \cdot 26 \\ \\ \]\[=52(25)(25+1)-\frac{25(25+1)(2 \cdot 25+1)}{3}+26^2 \\ =23426\]

  12. mathmath333
    • one year ago
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    long way

  13. freckles
    • one year ago
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    lol i did say a bit longer

  14. imqwerty
    • one year ago
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    :)

  15. freckles
    • one year ago
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    here is an example if what I did using a smaller exponent \[(a+b+c+d)^{3} \\ =([a+b]+[c+d])^3 \\ =(a+b)^3+c_1(a+b)^2(c+d)+c_2(a+b)(c+d)^2+(c+d)^3 \\ \text{ number of terms } =4+3(2)+2(3)+4\]

  16. freckles
    • one year ago
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    (a+b)^3 <--expanding this will give you 3+1 (a+b)^2(c+d) <--expanding this will give you well (3 terms)(c+d) which is 3+3 or 3(2) terms and so on...

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