Pagen13
  • Pagen13
Determine the most efficient way to use the Binomial Theorem to show the following. (11)^4=14641 a.write 11-5+6 and expand. b.write 11=10+1 and expand. c.write 11=3+3+2 and expand. d.write 11=4+4+3 and expand.
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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Pagen13
  • Pagen13
@JoeDeWise
anonymous
  • anonymous
here
Pagen13
  • Pagen13
Okay! I posted my question

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More answers

anonymous
  • anonymous
114 =11*11*11*11 =114 =14641
Pagen13
  • Pagen13
So?
freckles
  • freckles
I would prefer myself to evaluate powers of 10 or powers of 1 because they are extremely easy.
Pagen13
  • Pagen13
What do you mean?
freckles
  • freckles
for example which one seems easier? 5^4? 10^4? also c doesn't make sense because 3+3+2 is totally not 11 also 11=4+4+3 seems like it might be harder to use binomial theorem since 4+4+3 consist of 3 terms. But I guess you do binomial again and again to expand (4+4+3)^4.
freckles
  • freckles
what do I mean about what?
Pagen13
  • Pagen13
10^4 seems easier
freckles
  • freckles
right
freckles
  • freckles
10^4=10000 but 5^4 I would have to go back to multiplying 5*5*5*5 to figure out
Pagen13
  • Pagen13
Would it be B?
freckles
  • freckles
yes in my opinion expanding (10+1)^4 using binomial theorem is tons easier because like I said before evaluating powers of 10 and evaluating powers of 1 are way easy
freckles
  • freckles
\[(a+b)^4=a^4+4 \cdot a^3 b+6 \cdot a^2b^2 + 4 \cdot ab^3+b^4 \\ \text{ so } (10+1)^4=10^4+4 \cdot 10^3(1)+6 \cdot 10^2(1)^2+4 \cdot 10(1)^3+1^4\] try evaluating to see how easy that is
Pagen13
  • Pagen13
It was B.! Thank you! Could you help me with three more?
freckles
  • freckles
I can try
Pagen13
  • Pagen13
Use Pascal's Triangle to expand the binomial (d-3)^6
freckles
  • freckles
well using pascal's triangle what are the coefficients in the (6+1)th row?
Pagen13
  • Pagen13
I am unsure
freckles
  • freckles
so you don't know where the 7th row is?
Pagen13
  • Pagen13
I don't understand Pascal's Triangle..
freckles
  • freckles
|dw:1449683889849:dw| here I wrote 6 rows
freckles
  • freckles
the first row just is 1 the second row has 1 1 third row has 1 2 1 the fourth row has 1 4 6 4 1 a row is in a horizontal position, not vertical or diagonal
freckles
  • freckles
\[(a+b)^0=1 \\ (a+b)^1=a+b\\ (a+b)^2=a^2+2ab+b^2 \\ (a+b)^3=a^3+3a^2b+3ab^2+b^3 \\ (a+b)^4=a^4+4 a^3b+6a^2b^2+4ab^3+b^4 \\ (a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5 \] notice pascal's triangle gives you the coefficients of these expansions
freckles
  • freckles
and yes I could have but the 1's there but 1*a=a and 1*b=b and so on... \[(a+b)^0=1 \\ (a+b)^1=1a+1b\\ (a+b)^2=1a^2+2ab+1b^2 \\ (a+b)^3=1a^3+3a^2b+3ab^2+1b^3 \\ (a+b)^4=1a^4+4 a^3b+6a^2b^2+4ab^3+1b^4 \\ (a+b)^5=1a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+1b^5 \]
freckles
  • freckles
also notice in the expansions above the powers of the a is decreasing while the powers of the b is increasing also notice that the degree of each term is equal to what we are raising the binomial to
freckles
  • freckles
so can you expand (a+b)^6
Pagen13
  • Pagen13
(a+b)=a1 (a+b)=a1b and so on?
freckles
  • freckles
no a+b is not a1 and a+b is not a1b a+b is a+b or 1a+1b
Pagen13
  • Pagen13
Ugh..
freckles
  • freckles
I don't understand why you are saying a+b is equal to two different things...
Pagen13
  • Pagen13
I am just fully confused..
freckles
  • freckles
let's forget about the coefficients for a sec let's focus on the powers of the term... I will call the coefficients c0,c1,...blah blah\[(a+b)^8 \\ =a^8b^0+c_0 a^7b^1+c_1a^6 b^2+c_2 a^5b^3+c_3a^4b^4+c_4a^3b^5+c_5a^2b^6+c_6a^1b^7+a^0b^8\] look at this expansion sorta ignoring the constant values c that we can find off the pascal's triangle do you see what is happening with the powers here? do you see that each term has degree 8?
freckles
  • freckles
we don't need to actually write b^0 since this is 1 or a^0 sine this is also 1 we also normally write a^1 as just a and b^1 as just b I wrote the 1's and 0's because I thought it would be easier for you to recognize the pattern
freckles
  • freckles
look at the powers of a.... they are decreasing by 1 each time while the powers of b are increasing each time by 1 the degree of each term is 8 like for example 8+0=8 and 7+1=8 6+2=8 5+3=8 4+4=8 2+6=8 3+5=8 1+7=8 0+8=8
freckles
  • freckles
can you try expanding \[(a+b)^6\] you can put the c's in as I did above as placeholders for the actual values that we can find from the pascal triangle

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