blankasap
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#1
Report Thread starter 4 weeks ago
#1
find the constant term in (2x - 1/x^2) ^12
I used the coeffient rule to get the xs =x^0
(x)^12-r (x)^-2r= x^0

I got r=4
this is wrong pls helppp
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JAH36
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#2
Report 4 weeks ago
#2
https://www.doubtnut.com/question-an...-is--642904431
This may be helpful?
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davros
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#3
Report 4 weeks ago
#3
(Original post by blankasap)
find the constant term in (2x - 1/x^2) ^12
I used the coeffient rule to get the xs =x^0
(x)^12-r (x)^-2r= x^0

I got r=4
this is wrong pls helppp
r= 4 is correct for the power of 1/x^2 because that gives you a 1/x^8 which cancels with the x^8 (and 4 + 8 = 12). But you need the actual number that appears in the expansion which will be some power of 2 multiplied by some power of -1 multiplied by a binomial coefficient which you need to look up or work out
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blankasap
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#4
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#4
(Original post by davros)
r= 4 is correct for the power of 1/x^2 because that gives you a 1/x^8 which cancels with the x^8 (and 4 + 8 = 12). But you need the actual number that appears in the expansion which will be some power of 2 multiplied by some power of -1 multiplied by a binomial coefficient which you need to look up or work out
i'm sorry but i still don't get it :/
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davros
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#5
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#5
(Original post by blankasap)
i'm sorry but i still don't get it :/
You need to post your working so that someone can check it

Write down what the binomial formula is for expanding (a + b)^n. where n is an integer.

What is 'a' in this case?
What is 'b' in this case?
What is 'n' in this case?

Which (r th) term do you need to find, what are the powers and coefficients involved?
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blankasap
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#6
Report Thread starter 3 weeks ago
#6
(Original post by davros)
You need to post your working so that someone can check it

Write down what the binomial formula is for expanding (a + b)^n. where n is an integer.

What is 'a' in this case?
What is 'b' in this case?
What is 'n' in this case?

Which (r th) term do you need to find, what are the powers and coefficients involved?
The n is 12, the a is 2x and the b is -x^-2,
using the binomial expansion to = x^0
{2} (2)^12-r (x)^12-r (-1)^r (x^-2)^r
{ r}
Last edited by blankasap; 3 weeks ago
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davros
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#7
Report 3 weeks ago
#7
(Original post by blankasap)
The n is 12, the a is 2x and the b is -x^-2,
using the binomial expansion to = x^0
{2} (2)^12-r (x)^12-r (-1)^r (x^-2)^r
{ r}
OK, so you want the constant term i.e. with the power of x = 0. What equation can you form that involves 'r' that makes the power of x equal to 0? What are all the numerical parts multiplied together - i,e, binomial coefficient, power of 2, power of -1 etc?
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blankasap
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#8
Report Thread starter 3 weeks ago
#8
(Original post by davros)
OK, so you want the constant term i.e. with the power of x = 0. What equation can you form that involves 'r' that makes the power of x equal to 0? What are all the numerical parts multiplied together - i,e, binomial coefficient, power of 2, power of -1 etc?
oh i get it now, thank you
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davros
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#9
Report 3 weeks ago
#9
(Original post by blankasap)
oh i get it now, thank you
well done!
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