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LoginA maths (urgent) - binomial theorem
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given that ( 1 + kx)^n = 1-28x + 364x^2 +..., find the values of k and n. By expanding, I somehow got to know that nk = -28 and nC2(k)^2 is 364.
I dont know how to derive the correct answer but by trial and error i got n = 14 and k = -2. whats the proper way of doing this question?
the next question. in the expansion of ( 1 + x)^n, coefficient of x^4 and x^5 are the equal. find value of n.
Again by trial and error i got n = 9. but i do not know how to get the answer through proper presentation.
i wasnt in class today and my friend told me that some of the questions require some "new formula" besides the basic binomial statements. can someone help me identify whether these 2 questions are the ones that require the special formula or whatever it is.
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given that ( 1 + kx)^n = 1-28x + 364x^2 +..., find the values of k and n
Expand ( 1 + kx)^n gives you
1 + nC1 (kx) + nC2 (kx)^2 + ....
where nC1 = n and nC2 = n(n-1)/2so you get
( 1 + kx)^n = 1 + nkx + (n(n-1)k^2)/2 x^2 + ...resolving, nk = -28
(n(n-1)k^2)/2 = 364 -------> What you got so farfrom nk=-28, k=-28/n
so n(n-1)(-28/n)^2 = 364 * 2
784(n^2-n) = 728n^2
56n^2 - 784n = 0
n(n-14) = 0
n = 0 (reject) or n = 14thus, k=-2
the next question. in the expansion of ( 1 + x)^n, coefficient of x^4 and x^5 are the equal. find value of n.
This is somewhat similar
( 1 + x)^n = 1 + nx + nC2 x^2 + nC3 x^3 + nC4 x^4 + nC5 x^5 + ...
so nC4 = nC5
n(n-1)(n-2)(n-3) / {1*2*3*4} = n(n-1)(n-2)(n-3)(n-4) / {1*2*3*4*5}
120n(n-1)(n-2)(n-3) = 24n(n-1)(n-2)(n-3)(n-4)Bring over to one side
n(n-1)(n-2)(n-3) { 24(n-4) - 120 } = 0
know that n cannot be 0, 1, 2, or 3 because we have coefficients for x^4 and x^5, assuming the coefficients are non-zero.
so 24n -96-120 = 0
n = 9Add on: n=0,1,2,3 can be solutions if the coefficients of x^4 and x^5 are zero, which is why I said assuming the coefficients are non-zero. This is because in expansions up to n=3, we have till (1+x)^n = 1 + nx + ... + nx^2 + x^3 + 0x^4 + 0x^5 +0x^6 +....
Coefficients of x^n where n>4 is zero. But this is not what we are looking for.
Another way to solve the above question is to look at the Pascal Table
note for n=3, you get 1 3 3 1 => the 2nd and 3rd term is the same
similarly for n=5, you get 1 5 10 10 5 1 => the 3rd and 4th term is the same
for n=7, you get 1 7 21 35 35 21 7 1 => 4th and 5th term is the same
finally, for n=9, you get 1 9 36 84 126 126 84 36 9 1 => 5th and 6th term the same
and the 5th term is of x^4 and 6th term is of x^5.....Eagle
Owner of Strategic TuitionEdited by eagle 31 Jul `08, 9:43AM
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