Hi,
I have a couple more questions on Binomial Theorem.
I am not sure how to attach pictures to sgforums, so heres a link instead.
https://picasaweb.google.com/102868029261377164402/H2Math_BinomialTheorem?feat=directlink
Thank you for your assistance and time.
I really appreciate it. =)
Cheers,
Jennifer
Hi,
One should recall standard expansions like:
(1) 1 / (1 - x) = 1 + x + x^2 + x^3 + ... + x^n + ...,
of which the coefficient of x^n is 1.
Note that result (1) follows from the infinite sum of a geometric progression.
(2) 1 / (1 - x)^2 = 1 + 2x + 3x^2 + 4x^3 + ... + (n + 1)x^n + ...,
of which the coefficient of x^n is (n + 1).
(3) 1 / (1 - x)^3 = (1/2) [ (1)(2) + (2)(3) x + (3)(4) x^2 + ... + (n + 1)(n + 2) x^n + ... ],
of which the coefficient of x^n is (1/2) (n + 1)(n + 2).
In fact, result (2) can be obtained by differentiating result (1) and result (3) can be reached from result (2) in the same way.
Differentiation (knowledge from 'O' level) is an easier approach than the tedious general form involving factorials.
For 1 / (1 + x), 1 / (1 + x)^2 and 1 / (1 + x)^3, we need to add (-1)^n in front.
P.S. The syllabus does not mention explicitly the coefficient of the general term. It is also highly unlikely to be asked, since the last question on it appeared in 1991. The present trend is for binomial expansion to be asked along with maclaurin's expansion.
Thanks.
Cheers,
Wen Shih
Hi,
One way to find the appropriate value to substitute for x is to solve an equation, say,
(x + 2) / x = 5.5.
One must make sure that the value is within the validity range.
Another approach is by means of trial and error (which Cambridge never sets but instead gives the value directly). One practical choice of x is 1/k, where k is some integer.
For (1 + 4x)^(1/2) with validity range |x| < 1/4, we could consider
(a) x = 1/5 to give sqrt(5) an approximate value of 375/169
or
(b) x = -1/5 to give sqrt(5) an approximate value of 125/61.
(a) is better (than (b)) since it is closer to the decimal value of sqrt(5).
I feel that the question your school has set is not well-crafted, because it is not easy (for one without experience and numerical sense) to arrive at those (obscure) fractions.
Thanks.
Cheers,
Wen Shih
Hi,
Given an expression, the size of x determines how it will be expanded.
Consider (1 - x)^(-3) and x is small (rather than being close to zero, it's just a value within some validity range), then we expand it as usual.
If x is large, then we need to carry out some manipulation:
(1 - x)^(-3) = (-x)^(-3) (1 - 1/x)^(-3)
and then expand as usual, since 1/x is now small.
Thanks.
Cheers,
Wen Shih
Hi Mr Wee,
I am very impressed by the advice and tips you gave.
Instead of using factorials, I re-do these questions using the formula method (where u differentiate (1+x)^-1 and (1-x)^-1) and it was so much easier and faster.
You have certainly helped to clarify my doubts!
Thank you very much!
Best wishes,
Jennifer
Dear Jennifer,
You are welcome :)
Have a great weekend!
Cheers,
Wen Shih