### Analysis: Quiz 4

Analysis: Quiz 4

$1(2!)+(1+3)(3!)+(1+3+5)(4!)+(1+3+5+7)(5!)+\cdots+n\text{th term}$

Question 1:
Find the $n$th term for the series above.

A. $(1+3+5+...+n\text{th term})(n+1)!$
B. $(2n-1)(n+1)!$
C. $n^2(n+1)!$
D. None of the above

When we're asked to find the $n$th term for any given series, it's always an expression (or should I say, a formula) written in terms of $n$.

In solving problem like this, we need to be very observant and be able to recognize instantly the other possible representation that the current expression might be replaced with.

Having said that, we see that if we sum up $1,\,(1+3),\,(1+3+5),\,(1+3+5+7),\,\cdots$ we get $1,\,4,\,9,\,16,\,\cdots$ and that simply is the square of natural number starting from 1, i.e. $1^2,\,2^2,\,3^2,\,4^2+\cdots$

Therefore, we can rewrite the given series as

$1^2(2!)+2^2(3!)+3^2(4!)+4^2(5!)+\cdots+n\text{th term}$

Now we do see the pattern, the $n\text{th term}$ is actually a product of two factors, where its first factor is purely $n^2$ and the second factor is the factorial of the nth term from a sequence $2,\,3,\,4,\,\cdots$, i.e. $(2+(n-1)(1))!=(n+1)!$.

Combining both, we can say the $n$th term for the series $1(2!)+(1+3)(3!)+(1+3+5)(4!)+(1+3+5+7)(5!)+\cdots+n\text{th term}$ is $n^2(n+1)!$.

Therefore the correct answer is C.

Question 2:
What could be the possible sum of the given expression above?
A. $\infty$
B. A product of two factors
C. A sum/difference of two terms

In most cases, the sum of a given series could be simplified if the series can be expressed as a telescoping series. By definition, a telescoping series is a series whose partial sums eventually only have a fixed number of terms after cancellation. Therefore, the end result must also be sum/difference of two (or more) terms.
As always, words won't do it justice. I will hence show with one example why this should make sense to you.

A telescoping series

$=(a_1-a_2)+(a_2-a_3)+(a_3-a_4)+\cdots+(a_{n-2}-a_{n-1})+(a_{n-1}-a_n)$

$=(a_1-\cancel{a_2})+(\cancel{a_2}-\cancel{a_3})+(\cancel{a_3}-\cancel{a_4})+\cdots+(\cancel{a_{n-2}}-\cancel{a_{n-1}})+(\cancel{a_{n-1}}-a_n)$

$=a_1-a_n$

I am sure you can picture this now and therefore the correct answer is C.

Question 3:
Which aspect should we consider when finding the sum of the given function?
A. Arithmetic Series
B. Geometric Series
C. Binomial Sum
D. Telescoping Sum

As mentioned before, most problems that involved the sum with factorial are telescoping series. Hence, we will use telescoping method to deal with this particular problem and therefore the correct answer is D.

Question 4:
The problem solving method based on the answer you picked for question 3 assigns for_ _ _
A. Only one way of tackling the problem.
B. More than one way of tackling the problem.
We first must understand the basic in solving any given telescoping sum.

In general, we use the method of differences in our working such that:

If $u_r=f(r)-f(r-1)$,

then

[MATH]\sum_{r=1}^{n} u_r=\sum_{r=1}^{n}[f(r)-f(r-1)][/MATH]

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=f(1)-f(0)+f(2)-f(1)+f(3)-f(2)+\cdots+f(n-1)-f(n-2)+f(n)-f(n-1)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-f(0)+f(n)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=f(n)-f(0)$

Back to our problem, we have:

$1(2!)+(1+3)(3!)+(1+3+5)(4!)+(1+3+5+7)(5!)+\cdots+n\text{th term}$

[MATH]=\sum_{r=1}^{n} u_r[/MATH] (note that our $u_r$ is the $n$th term that we found from question (1))

[MATH]=\sum_{r=1}^{n} r^2(r+1)![/MATH]

If we could rewrite $r^2(r+1)!$ as the difference of two functions in the form $f(r)-f(r-1)$, then we would solve the problem.

The question is, in how many ways exactly there have to express  $r^2(r+1)!$ as the difference of two functions that takes the form $f(r)-f(r-1)$? That is what being asked in this question (4).

Regardless if we're superbly powerful in exploiting the given algebraic expression, there is always only ONE way of rewriting $r^2(r+1)!$ in the form $f(r)-f(r-1)$:

$r^2(r+1)!$

$=(r+1)!(r^2)$ we know we need to introduce a difference of two functions here

$=(r+1)!(\text{something}-\text{something else})$ we should bring in $r+2$ so we get $(r+2)!$

$=(r+1)!((r+2)(r-1)-(r-2))$

$=(r+1)!(r+2)(r-1)-(r-2)(r+1)!$

$=(r+1)!(r+2)(r-1)-(r+1)!(r-2)$

$=(r+2)!(r-1)-((r-1)+2)!((r-1)-1)$

$=f(r)-f(r-1)$

[MATH]\therefore \sum_{r=1}^{n} r^2(r+1)![/MATH]

$=(r+2)!(r-1)-((r-1)+2)!((r-1)-1)$

$=f(r)-f(r-1)$

$=f(n)-f(0)$ here we will replace $r=n$ and $r=1$ respectively into $f(r)$ and $f(r-1)$

$=(n+2)!(n-1)-((1-1)+2)!((1-1)-1)$

$=(n+2)!(n-1)-(2)!(-1)$

$=(n+2)!(n-1)+2$

Therefore the answer for question (4) is A and the answer for question (5) is $(n+2)!(n-1)+2$.