Binomial Theorum Questions (1 Viewer)

[GaDaMIt]

A few questions, only just started this topic so im really lost

Without expanding, simplify
a) 1 + 3(x-1) + 3(x-1)^2+(x-1)^3

This question assumes no knowledge of "C" as it is in chapter 5A of the cambridge book, much before anything gets difficult but im already confused

Similar question in next exercise but now knowledge of C is assumed i believe..

y^5 + 5y^4(x-y) + 10y^3(x-y)^2 + 10y^2(x-y)^3 + 5y(x-4)^4 + (x-y)^5


{[(sqrt3 + 1)^4] + [(sqrt3 - 1)^4]} divided by
{[(sqrt3 + 1)^4][(sqrt 3 - 1)^4]}

Note that just says sqrt 3, not the 1s.

Im at a loss as to how to simplify both the numerator and the denominator, please explain for both?


Thanks for any help..

 

[GaDaMIt]

OK - first one you should recognise that as the expansion of a cubic - ie in the form (a + b)³ = a³ + 3a²b + 3ab² + b³
In this case
a = 1
b = x - 1
(a + b)³ = x³

Simlar for the next one except it's to the fifth - that requires you to know the expansion of (x + y)ⁿ which it sounds like you haven't learnt yet.

I do know the expansion, its just whenever i get a new topic in maths i never know how to set out my working, or i cant judge what type of working is required...



Also to others .. sorry i bolded 2 of the questions and not the last, but yeh, if i could get a reply to the last one i would be very appreciative

 

[SoulSearcher]

Expansion of (rt3 +1)⁴
= (rt3)⁴ + 4(rt3)³ + 6(rt3)² + 4(rt3) + 1
= 28 + 16rt3
Expansion of (rt3 -1)⁴
= (rt3)⁴ - 4(rt3)³ + 6(rt3)² - 4(rt3) + 1
= 28 - 16rt3

Therefore (rt3 +1)⁴ + (rt3 -1)⁴
= (rt3)⁴ + 4(rt3)³ + 6(rt3)² + 4(rt3) + 1 + (rt3)⁴ - 4(rt3)³ + 6(rt3)² - 4(rt3) + 1
= 2[(rt3)⁴ + 6(rt3)² + 1]
= 56

And (rt3 +1)⁴(rt3 -1)⁴
= (28+16rt3)(28-16rt3)
= 784-768
= 16

Therefore [(rt3 +1)⁴+(rt3 -1)⁴] / [(rt3 +1)⁴(rt3 -1)⁴] = 56/16 = 3.5

 

[GaDaMIt]

One more question.. when you're asked to find the middle term in a sequence, say (2x - 3y)^10.. how would you go about it? I just figured from the general term that the middle term would have equal powers for x and y, thus 10 - k = k .. 2k = 10, k = 5, but just wanna make sure on this


And also, what difference is there between

Tk = nCk . a^(n - k) . b^k

and

T(k+1) = nCk . a^(n - k) . b^k

where as is evident, only the term changes. Can both of these be applied to questions? If so, different terms would be yeilding the same answer?? If not, well I was told earlier by my friend that he prefers use Tk one..., and teacher taught us T(k+1)...

 

[webby234]

Yes - middle term k will equal n - k, in your case it will be the 6th term, when k = 5

And the T_k one won't get you the right answer in certain questions ie when it asks for the 6th term if you just subbed 6 in for k you would get the wrong answer. You need to use T_k+1 simply because the first term T₁ is when k = 0, for the second term k = 1 etc.

 

[GaDaMIt]

lol