Binomial Theorum (1 Viewer)

[king_of_boredom]

what is the greatest coefficient in (x + 2)^8?

thx in adv.

(if possible, can you please solve using T_k+1 / T_k = n - k + 1/k . b/a? thanks).

 

[haboozin]

what is the greatest coefficient in (x + 2)^8?

thx in adv.

(if possible, can you please solve using T_k+1 / T_k = n - k + 1/k . b/a? thanks).

k = 5 or 6


greatest coeff = 1792



2 * 8!/(K + 1)!(7-k)! * k!(8-k)!/8! > 1

simplifiy....

(16 - 2k)/(k+1) > 1

simplify
k < 5

therefore k = 5 or 6

 

[Alexluby]

(x+2)^8

T_(r+1) = 8 C_r x^(8-r). 2^r
T_r = 8 C_(r-1) x^(9-r) . 2^(r-1)

Let C_r, and C_(r-1) be the coefficients of the terms....

then [C_(r+1)]/ C_r = [8 C-r 2^r] / 8 C_(r-1) 2^(r-1)
= (18-2r)/r

C_(r+1) >= C_r if 18 - 2r >= r
18 >= 3r
r <= 5
Then you'll do the proving that when r = 5 and so on...

Greatest coefficient : is C_6 = 8 C_5 1^3 . 2^5
= 56 * 32
= 1792