3U Binomial crazyness (1 Viewer)

[underthesun]

not so long ago someone posted a 3u question here in hoping to get more views. I reckon we should rename the forums?

anyways here goes:

find n if the coefficients of the 2nd, 3rd and 4th terms in the expansion of (1 + x)ⁿ are the successive terms of an Arithmetic Sequence.

my working,

ⁿC₂ - ⁿC₁ = ⁿC₃ - ⁿC₂

from there, just change to the real formula (n! / (n-k)!k!) stuff, and i got a quadratic with no roots for n..

it's :: n² - 9n + 14 = 0

try it yourself, it doesn't take long, but i just want to confirm that the question is wrong..

 

[wogboy]

it's :: n2 - 9n + 14 = 0

Actually that quadratic has two integer roots, n=2 and n=7 (use the quadratic formula or factorise it).

n^2 - 9n + 14 = (n-2)(n-7)

Since 2C3 is undefined, the answer must be n=7. If you try testing this answer on a calculator, you'll see that it's true.

 

[underthesun]

whoops..