Binomial Expansion (1 Viewer)

klee98 · Feb 7, 2015

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

klee98 · Feb 9, 2015

InteGrand · Feb 9, 2015

klee98 said:
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 $S_n \equiv \sum_{k=1}^{n}\frac{k}{(k+1)!}=\frac{(n+1)!-1}{(n+1)!}$ .

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

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

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

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

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

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

Binomial Expansion (1 Viewer)

klee98

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