Binomial Expansion (1 Viewer)

klee98

Member
I've changed the question!
I know how to do a,b and d
How do you do c?
Any help is greatly appreciated

Attachments

• 43.7 KB Views: 51
Last edited:

._.

InteGrand

Well-Known Member
I've changed the question!
I know how to do a,b and d
How do you do c?
Any help is greatly appreciated

If you evaluate the partial sums from part (b), you'll see that the guess to use is $\bg_white S_n \equiv \sum_{k=1}^{n}\frac{k}{(k+1)!}=\frac{(n+1)!-1}{(n+1)!}$.

I'll only show the induction step. (I assume you can easily do the base case step, assuming you know how to do mathematical induction proofs.)

Assuming $\bg_white S_{n-1} =\frac{n! - 1}{n!}$, we show $\bg_white S_{n} =\frac{(n+1)!-1}{(n+1)!}$.
.

$\bg_white S_n = S_{n-1}+T_n$, where $\bg_white T_n$ is the nth term of the series in the question (this is the typical way to do summation induction proofs)

$\bg_white = \frac{n!-1}{n!} + \frac{n}{(n+1)!}$, by the inductive hypothesis and the fact that $\bg_white T_k = \frac{k}{(k+1)!}$ (given)

$\bg_white = \frac{(n+1)(n!-1)+n}{(n+1)!}$ (using factorial rules to take a common denominator, the rule being (n+1)! = (n+1).n!)

$\bg_white =\frac{(n+1)!-(n+1)+n}{(n+1)!}$ (expanding the numerator)

$\bg_white =\frac{(n+1)!-1}{(n+1)!}$ (simplifying the numerator), which is what we wanted to show. $\bg_white \blacksquare$