Want to help us with this year's BoS Trials? Let us know before 30 June. See this thread for details
Looking for HSC notes and resources? Check out our Notes & Resources page

binomial expansion (1 Viewer)

Thread starter Hello_World2
Start date Oct 7, 2022

H

Hello_World2

Member

#1

I've done part a, but I am not sure how to use it to get parts b and c. Help would be appreciated, thanks!

Screen Shot 2022-10-07 at 3.02.29 pm.png

C

cossine

Well-Known Member

#2

Hello_World2 said:
I've done part a, but I am not sure how to use it to get parts b and c. Help would be appreciated, thanks!
View attachment 36477

Try mathematical induction.

Hamburgler

Member

#3

Consider expansion of

Hamburgler

Member

#4

As for part C consider integrating both sides in

And letting x=-1

H

hogzillaAnson

Member

#5

$\begin{align*} \\&(c)\ \text{By equating the expression for}\ f(x)\ \text{in}\ (b)\ \text{with the expression given at the start, we know that}\\ &\sum_{r=1}^{n} \binom{n}{r} x^{r-1} = \sum_{r=1}^{n} (1+x)^{r-1} \\ &\text{We can now integrate both sides of the expression from} -1\ \text{to}\ 0\text{, giving}^{*}\\ &\sum_{r=1}^{n} \int_{-1}^{0}\binom{n}{r} x^{r-1}dx = \sum_{r=1}^{n} \int_{-1}^{0} (1+x)^{r-1} dx \\ &\sum_{r=1}^{n} \left[ \frac{x^r}{r}\binom{n}{r}\right]_{-1}^{0} = \sum_{r=1}^{n} \left[\frac{(1+x)^r}{r}\right]_{-1}^{0} \\ &\sum_{r=1}^{n} \left( 0-\frac{(-1)^r}{r}\binom{n}{r} \right) = \sum_{r=1}^{n} \left(\frac{1^{r}}{r} - 0 \right) \\ &\sum_{r=1}^{n} \frac{(-1)^{r+1}}{r}\binom{n}{r} = \sum_{r=1}^{n} \frac{1}{r} \\ &^{*} \int_a^{b}(f(x) + g(x))dx = \int_a^{b} f(x) dx + \int_a^{b} g(x) dx\ \text{, so we can reverse the order of integration}\ \smallint\ \text{and summation}\ \Sigma \end{align*}$

You must log in or register to reply here.

Share:

Facebook Twitter Reddit Pinterest Tumblr WhatsApp Email Link

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top