underthesun
N1NJ4
- Joined
- Aug 10, 2002
- Messages
- 1,781
- Location
- At the top of Riovanes Castle
- Gender
- Undisclosed
- HSC
- 2010
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)<sup>n</sup> are the successive terms of an Arithmetic Sequence.
my working,
<sup>n</sup>C<sub>2</sub> - <sup>n</sup>C<sub>1</sub> = <sup>n</sup>C<sub>3</sub> - <sup>n</sup>C<sub>2</sub>
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 :: n<sup>2</sup> - 9n + 14 = 0
try it yourself, it doesn't take long, but i just want to confirm that the question is wrong..
anyways here goes:
find n if the coefficients of the 2nd, 3rd and 4th terms in the expansion of (1 + x)<sup>n</sup> are the successive terms of an Arithmetic Sequence.
my working,
<sup>n</sup>C<sub>2</sub> - <sup>n</sup>C<sub>1</sub> = <sup>n</sup>C<sub>3</sub> - <sup>n</sup>C<sub>2</sub>
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 :: n<sup>2</sup> - 9n + 14 = 0
try it yourself, it doesn't take long, but i just want to confirm that the question is wrong..
