Bionomial thoerem

**raza9990** · February 4th 2010, 12:04 PM

write down and simplify the expression of (1-p)^5 .Use this result to find the expansion of (1-x-x^2)^5 in ascending powers of x as far as the terms in x^3 . Find the value of x which would enable you to estimate (0.9899)^5 from this expansion

**e^(i*pi)** · February 4th 2010, 01:02 PM

Originally Posted by raza9990

write down and simplify the expression of (1-p)^5 .Use this result to find the expansion of (1-x-x^2)^5 in ascending powers of x as far as the terms in x^3 . Find the value of x which would enable you to estimate (0.9899)^5 from this expansion

$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ... + \frac{n!}{n!}x^n$

In your case it will be a finite sequence.

**raza9990** · February 4th 2010, 07:43 PM

Originally Posted by e^(i*pi)

$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ... + \frac{n!}{n!}x^n$

In your case it will be a finite sequence.

but the question isn't solved yet . I tried it.
can you write up the solution steps for me .
I'm still confused .

**e^(i*pi)** · February 5th 2010, 06:04 AM

Originally Posted by raza9990

write down and simplify the expression of (1-p)^5 .Use this result to find the expansion of (1-x-x^2)^5 in ascending powers of x as far as the terms in x^3 . Find the value of x which would enable you to estimate (0.9899)^5 from this expansion

I'll expand the first part for you and give guidance on the others but I'm not doing your work for you in full.

$<br /> (1-p)^5 = 1 - 5p + \frac{5 \cdot 4}{2!}p^2 - \frac{5 \cdot 4 \cdot 3}{3!}p^3 + \frac{5 \cdot 4 \cdot 3 \cdot 2}{4!}p^4 - \frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{5!}p^5$

Any further terms will have multiplication by 0 so there are no more terms. I'll leave simplifying that to you.

What the next question is asking is to substitute $(x+x^2)$ for p in the above expansion. You can either do this as above or substitute $p = x(1+x)$ , if you use the latter remember that the whole expression is squared.

For the third part it is asking you to solve $1-q = 0.9899$ and $q = x^2+x$ where q is a constant.

**raza9990** · February 5th 2010, 10:39 AM

Originally Posted by e^(i*pi)

I'll expand the first part for you and give guidance on the others but I'm not doing your work for you in full.

$<br /> (1-p)^5 = 1 - 5p + \frac{5 \cdot 4}{2!}p^2 - \frac{5 \cdot 4 \cdot 3}{3!}p^3 + \frac{5 \cdot 4 \cdot 3 \cdot 2}{4!}p^4 - \frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{5!}p^5$

Any further terms will have multiplication by 0 so there are no more terms. I'll leave simplifying that to you.

What the next question is asking is to substitute $(x+x^2)$ for p in the above expansion. You can either do this as above or substitute $p = x(1+x)$ , if you use the latter remember that the whole expression is squared.

For the third part it is asking you to solve $1-q = 0.9899$ and $q = x^2+x$ where q is a constant.

thanks ...you are a savior

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