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anonymous

  • 4 years ago

How do I expand (2g+2)^8

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  1. Shayaan_Mustafa
    • 4 years ago
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    do you know binomial theorem?

  2. anonymous
    • 4 years ago
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    no :(

  3. Shayaan_Mustafa
    • 4 years ago
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    hmmm... it is much long and complicated. Do you know what is factorial?

  4. Shayaan_Mustafa
    • 4 years ago
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    here is the link. and try to understand. i will surely help you. http://www.intmath.com/series-binomial-theorem/4-binomial-theorem.php

  5. anonymous
    • 4 years ago
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    thank you

  6. Shayaan_Mustafa
    • 4 years ago
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    welcome.

  7. anonymous
    • 4 years ago
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    If I were to expand (3-t)^3 is it 3^3+3(3^2-t)+3(3-t^2)-t^2

  8. Shayaan_Mustafa
    • 4 years ago
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    no... did you use it correctly?

  9. anonymous
    • 4 years ago
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    i think so. what did i do wrong?

  10. Shayaan_Mustafa
    • 4 years ago
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    here your brackets are wrong. they should look like this. 3(3^2)t and 3(3)t^2

  11. Shayaan_Mustafa
    • 4 years ago
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    do you want to learn basic of binomial theorem?

  12. anonymous
    • 4 years ago
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    sure :D

  13. Shayaan_Mustafa
    • 4 years ago
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    ok. it will take much time. let us start. do you know the formula (a+b)^2=a^2+2ab+b^2

  14. anonymous
    • 4 years ago
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    yea

  15. Shayaan_Mustafa
    • 4 years ago
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    good. it is also a binomial theorem. Notice that here power two indicates that when we expand our formula, we get one more term than the no. of exponent. Here exponent is 2 while when we expanded then we got 3 terms.

  16. anonymous
    • 4 years ago
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    ok

  17. Shayaan_Mustafa
    • 4 years ago
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    similarly, I hope you know the formula (a+b)^3=a^3+3a^2b+3ab^2+b^3 here exponent is 3 while expanded terms are 4. This implies that expanded terms will always be one more than the exponent. Now is it clear?

  18. anonymous
    • 4 years ago
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    oo I think I get it now. Thank you.

  19. Shayaan_Mustafa
    • 4 years ago
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    welcome. ok now make a step ahead. here is binomial formula. \[(a+b)^{n}=C _{0}^{n}a ^{n}+C _{1}^{n}a ^{n-1}b+C _{2}^{n}a ^{n-2}b ^{2}.....\] Where C stands for combination. in general this combination is written as, \[C _{r}^{n}=n!/(r!(n-r)!)\] now in above formula is constantly increasing, r=0,r=1,r=2.......

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