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)/2
so 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
far
from 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 = 14
thus, 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 =
9
Add 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 Tuition