C2 BINOMIAL EXPANSION HELP!! pleaseeeee
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C2 BINOMIAL EXPANSION HELP!! pleaseeeeeCan somebody show me how to solve such equations? i just cant get my head around this :-
1. The coefficinet of x^3 in the expansion of (3+bx)^5 is -720. Find the value of the constant b.
2. given that
(2+x)^5 (2-x)^5 = A+Bx^2+Cx^4, find the value of constants A,B and C.
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Re: C2 BINOMIAL EXPANSION HELP!! pleaseeeeeI use(Original post by sharon800)
Can somebody show me how to solve such equations? i just cant get my head around this :-
1. The coefficinet of x^3 in the expansion of (3+bx)^5 is -720. Find the value of the constant b.
2. given that
(2+x)^5 (2-x)^5 = A+Bx^2+Cx^4, find the value of constants A,B and C.
Thank you
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Re: C2 BINOMIAL EXPANSION HELP!! pleaseeeeeFor the first one you:
1. Expand the brackets until you get ......x^3
2. Make this coefficient equal to -720
3. Simple algebra from here
For the second one you:
1. expand both to give the answer
2. Total up the coeffients
3. From here, work out what A, B, C are -
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Re: C2 BINOMIAL EXPANSION HELP!! pleaseeeeeThank you(Original post by Frodo Baggins)
For the first one you:
1. Expand the brackets until you get ......x^3
2. Make this coefficient equal to -720
3. Simple algebra from here
For the second one you:
1. expand both to give the answer
2. Total up the coeffients
3. From here, work out what A, B, C are
for the first one
in the expansion of x^3 i got 5C3 *(3)^3*bx^2 = which is i think 270bx^3 = -720...and my answer is wrong...i dont know what i have done wrong -
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Re: C2 BINOMIAL EXPANSION HELP!! pleaseeeeeThis is(Original post by sharon800)
Thank you
for the first one
in the expansion of x^3 i got 5C3 *(3)^3*bx^2
and even if it were the correct term you have not squared your bLast edited by TenOfThem; 05-04-2012 at 10:19. -
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Re: C2 BINOMIAL EXPANSION HELP!! pleaseeeeeI got b as -2(Original post by sharon800)
Thank you
for the first one
in the expansion of x^3 i got 5C3 *(3)^3*bx^2 = which is i think 270bx^3 = -720...and my answer is wrong...i dont know what i have done wrong
Basically, to get x^3
its 5c3 * (3)^2 * b^3x^3
so 90b^3=-720
so b^3=-8
so b=-2
i am unclear as to why i have got a neg. Is it cos im clever?Last edited by Frodo Baggins; 05-04-2012 at 12:37. -
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Re: C2 BINOMIAL EXPANSION HELP!! pleaseeeeeThank you! i did this last nigh, and got it(Original post by Frodo Baggins)
I got b as -2
Basically, to get x^3
its 5c3 * (3)^2 * b^3x^3
so 90b^3=-720
so b^3=-8
so b=-2
i am unclear as to why i have got a neg. Is it cos im clever?
how do you do the second one? -
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Re: C2 BINOMIAL EXPANSION HELP!! pleaseeeee(Original post by sharon800)
Thank you! i did this last nigh, and got it
how do you do the second one?
Expand both
and 
by using binomial expansion, then multiply both and compare the coefficients to find the value of A, B and C.
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Re: C2 BINOMIAL EXPANSION HELP!! pleaseeeeei got A= 0? DOESNT sound right:/(Original post by raheem94)
Expand both and by using binomial expansion, then multiply both and compare the coefficients to find the value of A, B and C.
and the coefficents of b and c huge numbers? can you please solve this, so i can compare the answer? -
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Re: C2 BINOMIAL EXPANSION HELP!! pleaseeeeeI get A as 1024. Try again.(Original post by sharon800)
i got A= 0? DOESNT sound right:/
and the coefficents of b and c huge numbers? can you please solve this, so i can compare the answer? -
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Re: C2 BINOMIAL EXPANSION HELP!! pleaseeeeelol i got 3125...:/(Original post by raheem94)
I get A as 1024. Try again.
how do you do it?
i did 5C0*(5)^5*(-X)^0 and got -3125 -
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Re: C2 BINOMIAL EXPANSION HELP!! pleaseeeee(Original post by sharon800)
lol i got 3125...:/
how do you do it?
i did 5C0*(5)^5*(-X)^0 and got -3125
So
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Re: C2 BINOMIAL EXPANSION HELP!! pleaseeeeeNo problem, you are welcome.(Original post by sharon800)
thank you! i know where i went wrong...instead of putting 2^5 i put s^5 lol
i get it! thanks again!
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Re: C2 BINOMIAL EXPANSION HELP!! pleaseeeeeYes, it is not allowed to post full solutions.
Full solutions are considered a last resort.
Actually i didn't post the full solution, i only calculated one value, of A, to show the OP how to calculate them. He also had to calculate B and C, so it isn't a full solution, just a part of the solution to let the OP get started.
Why do you think it to be a full solution?
Have you read the OP's question?
![(2+ax)^4 = 2^4 (1+\frac{ax}{2})^4 = 2^4[1 + 4\frac{ax}{2} + \frac{4*3}{2!}(\frac{ax}{2})^2 + \frac{4*3*2}{3!}(\frac{ax}{2})^3 + (\frac{ax}{2})^4]](http://www.thestudentroom.co.uk/latexrender/pictures/71/7112ae6e9ef619e3106d441b35075262.png "(2+ax)^4 = 2^4 (1+\frac{ax}{2})^4 = 2^4[1 + 4\frac{ax}{2} + \frac{4*3}{2!}(\frac{ax}{2})^2 + \frac{4*3*2}{3!}(\frac{ax}{2})^3 + (\frac{ax}{2})^4]")
