highest coefficient (1 Viewer)

[gr_111]

find the highest coefficient of (2x-1)^5

I've tried the T_k+1 / T_k >1 and i'm getting k as -6 for some reason? Can anyone give a hint or another way to work it out, short of expanding the whole thing?

 

[RealiseNothing]

The fact that the second term is '-1' means that the 2nd, 4th, and 6th co-efficients will be negative whilst the 1st, 3rd, and 5th will be positive. Hence the highest co-efficient is either the 1st, 3rd, or 5th.

Now consider Pascal's triangle, for 6 terms: 1, 5, 10, 10, 5, 1.

Also consider the powers of 2 for the 1st, 3rd, and 5th terms:

1st term = 2^5
3rd term = 2^3
5th term = 2^1.

Hence it is obvious that the 3rd term is the highest co-efficient. Which is .

 

[gr_111]

Yep thats pretty clear now thanks. I thought i had these questions nailed but this one stuffed it. Why doesn't the T_k+1 / T_k >1 method work here?

 

[pokka]

When we find the greatest coefficient, we are generally concerned about the magnitude of the coefficients. So technically, it should be |T_k+1 / T_k| > 1. The reason for this, is because this inequality relies on the assumption that the coefficients increase up to T_k+1 and then decrease afterwards. i.e. T_1 < T_2 < ..... < T_k < T_k+1 > T_k+2 > ..... > T_n+1. In this example, the negative sign makes the coefficients alternate in sign so this pattern doesn't hold unless you ignore those negative signs.

 

[gr_111]

Yep ok that makes sense now. I just checked my working and realised i'd put 3^3 in my calc. instead of 2^3, so stupid mistake!

 

[gr_111]

Also, whats the chances of having no binomial in the HSC, like 2007? lol How could they just not test a whole topic?