if the coefficient of x,x*x,x*x*x in the binomial expansion (1/2+x) whole raise to 2n are in AP. Show that 2nsquare- 9n +7 is equal to zero?
by the binomial expansion:
(1/2 + x)^(2n)
= (1/2)^(2n) + (2n)(1/2)^(2n-1) x + 2n(2n - 1)/2! (1/2)^(2n-2) x^2 + 2n(2n-1)(2n-2)/3! (1/2)^(2n-3) x^3 + ...
so the coefficients of x, x^2, and x^3 are
2n(1/2)^(2n-1) n , 2n(2n-1)/2 (1/2)^(2n-2) and 2n(2n-1)(2n-2)/6 (1/2)^2n-3)
or
n (1/2)^(2n-2) , n(2n-1) (1/2)^(2n-2) and n(2n-1)(n-1)/3 (1/2)^(2n-2)
(sure hope I made no typo so far .....)
So these are supposed to be in AP
-- looks a bit messy, so.....
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Wow, let's work backwards, if they are in AP, then we are to show that
2n^2 - 9n + 7 = 0
(2n-7)(n-1) = 0
n = 7/2 or n = 1
if n=1, we would have had (1/2 + x)^2 which wouldn't even give us an x^3, so that's out!
if n = 7/2, then 2n would have been 7
and our expansion would have been
(1/2 + x)^7
giving us terms of
(1/2)^7 + 7(1/2)^6 x + 7(6)/2! (1/2)^5 x^2 + 7(6)(5)/3! (1/2)^4 x^3 + ..
so the coefficients of x, x^2 and x^3 would be
7/64 , 21/32 , and 35/16
or 7/64 , 42/64 and 140/64
all we have to do is test if these are in AP,
42/64 - 7/64 = 35/64
and 140/64 - 42/64 = 98/64 ≠ 35/64
so something is not right, either I made a mistake or the statement is false
(1/2 + x)^(2n)
= (1/2)^(2n) + (2n)(1/2)^(2n-1) x + 2n(2n - 1)/2! (1/2)^(2n-2) x^2 + 2n(2n-1)(2n-2)/3! (1/2)^(2n-3) x^3 + ...
= (1/2)^(2n-3) * (... + 8nx + 2n(2n-1)x^2 + 1/3 (2n)(2n-1)(2n-2)x^3 + ...)
So, we have an AP with terms
8n,2n(2n-1),4/3 n(2n-1)(n-1)
The difference must be constant, so we need
2(2n-1)-8n = 4/3 n(2n-1)(n-1) - 2n(2n-1)
3n(2n-1)-8n = n(2n-1)(n-1)
2n^3 - 9n^2 + 12n = 0
2n^2 - 9n + 12 = 0
no real roots
So, while my condition on n is different, it still appears that there is no solution. Better check my math as well.
Oops. I mad a mistake on line 1 of AP stuff:
2n(2n-1)-8n = 4/3 n(2n-1)(n-1) - 2n(2n-1)
4n(2n-1)-8n = n(2n-1)(n-1)
2n^2-11n+13 = 0
no rational roots
so, still no joy.