Question

Find a formula for partial sum of series
\[\large 1!(1^2+1)+2!(2^2+1)+3!(3^2+1)+\cdots+n!(n^2+1)\]

Answer 1

How are the processes of natural selection and artificial selection similar? How are they different? u help i help u

Answer 2

why r there random exclamation marks?

Answer 3

that exclamation mark has a different meaning in math, it is a short way for writing a product . For example, 3! and 10! mean this :
\[\begin{align}3! &= 3\times 2\times 1\\10!&=10\times 9\times 8 \times 7\times 6\times 5\times 4\times 3\times 2\times 1\end{align}\]

I'm pretty sure you will find answer to your biology question in your lesson pages... look it up!

Answer 4

Factoral sum is fun and damn complicated. See WolframAlpha section here: http://mathworld.wolfram.com/FactorialSums.html

The only one with a simple form is \[\sum_{k=1}^{n}k*k!\]
But the term in urs is k^2*k! and k^0*k! respectively. So I don't know. Sorry!

Answer 5

Ahh how do you work \(\sum\limits_{k=1}^{n}k*k!\) ?

Answer 6

\(\huge\bf\rlap{\color{Lime}{Thanks\ Good\ Bye}}{\color{Blue}{\; Thanks\ Good\ bye}}\)

Answer 7

why did that one look so weird :|

Answer 8

Wolfram said it is equal to (n+1)! -1 and it appears you can prove that by induction.

Answer 9

Yup - \[\sum_{k=1}^{n}k*k! = (n+1)! - 1\]
can indeed be proven by induction :)

Answer 10

Base case : \(\sum\limits_{k=1}^1=1=2-1=(1+1)!-1~~\color{green}{\checkmark}\)

Induction hypothesis : Let \(t\in\mathbb{N}\) be an integer greater than \(1\) and assume that the proposition is true for \(n=t\). That is \(\sum\limits_{k=1}^tk*k!=(t+1)!-1\)

Induction step :
\(\begin{align}\sum\limits_{k=1}^{t+1} k*k! &=t+1+\sum\limits_{k=1}^{t} k*k!\\~\\ &=t+1+ (t+1)!-1 \\~\\&=?\end{align}\)

how to conclude ?

Answer 11

First line on the induction step is missing some (k+1) terms there. Check again n u will see :)

Answer 12

Wow silly mistake! let me fix it thnks

Base case : \(\sum\limits_{k=1}^1=1=2-1=(1+1)!-1~~\color{green}{\checkmark}\)

Induction hypothesis : Let \(t\in\mathbb{N}\) be an integer greater than \(1\) and assume that the proposition is true for \(n=t\). That is \(\sum\limits_{k=1}^tk*k!=(t+1)!-1\)

Induction step :
\(\begin{align}\sum\limits_{k=1}^{t+1} k*k! &=(t+1)(t+1)!+\sum\limits_{k=1}^{t} k*k!\\~\\ &=(t+1)(t+1)!+ (t+1)!-1 \\~\\&=(t+1)!(t+1+1)-1\\~\\&=(t+2)!-1\end{align}\)

as desired \(\blacksquare\)

Answer 13

I have (t+1)! - 1 + (t +1)*(t+1)! in that line.

Answer 14

gotch! i think we can try something similar for the actual problem too

Answer 15

U ask n u will receive. Please put ur seat belt on....

Answer 16

but.. but.. don't we need to know the formula upfront for induction

Answer 17

Let G(n) = \[\sum_{k=1}^{n}(k^2+1)*k!\]

So G(n) + 2((n+1)!-1)
=\[\sum_{k=1}^{n}((k^2+1)*k! + 2k*k!)\]
=\[\sum_{k=1}^{n}(k^2+2k+1)*k!\]
=\[\sum_{k=1}^{n}(k+1)^2*k!\]
=\[\sum_{k=1}^{n}(k+1)*(k+1)!\]
= G(n+1) - G(1)

U should be able to go on from there :)

Fun one!

Answer 18

that looks brillinat!!!
so do we have
\[2((n+1)!-1)=G(n+1)-G(n)\]

which telescpes ?
let me try

Answer 19

In summary

G(n) + 2((n+1)! - 1) = G(n+1) - G(1);
since G(1)=2;
G(n+1) = G(n) + 2*(n+1)!

:)

Answer 20

\[G(n) + 2((n+1)!-1) = G(n+1)-G(1) \\~\\G(n+1)=G(n)+2(n+1)!\]

this looks like a recurrence relation ?

Answer 21

@rational

Yes indeed. dont know if u want to go on from here thou.

Answer 22

thats indeed a clever way to get a recurrence relation

Answer 23

but thats the dead end i guess as solving it looks painful

Answer 24

@rational agreed - did ur assignment ask for a specific solution?

been waiting to see @wio may have something fun to say here :P

Answer 25

me too, explicit formula is good... otherwise recurrence relation is the next best thing we can have

Answer 26

Also since this is a summation, there is a trivial recurrence relation always :

\[\sum\limits_{k=1}^{n+1}f(k) = f(n+1)+\sum\limits_{k=1}^{n}f(k) \]

So im guessing the question is really about finding an explicit formula..

Answer 27

I was trying a different recursion:\[
(n+1)!((n+1)^2+1) = n!(n^3+3n^2 +4n+2)\\
=n!(n(n^2+1) + 3(n^2+1) +(3n-1))\\
=(n+3)n!(n^2+1)+(3n-1)n!
\]\[
f(n+1) = (n+3)f(n)+ (3n-1)n!
\]When I tried to expand and get a pattern I got:\[
f(n+k) =(n+3)^kf(n)+(3n-1)n!\left(\frac{1-(n+3)^k}{1-(n+3)}\right)
\]I guess letting \(k=0\) would check whether the work is correct

Answer 28

Interesting.. letting k=0 gives f(n) = f(n), so it checks out

Answer 29

If there was a nice telescope, it would require that there be some nice \(k\) value, but I don't see it.

Answer 30

DAMN it - serve me right for jumping steps n not writing things out...mistake at the next to the last step :(

G(n) + 2*(n+1)! - 2
=\[\sum_{k=1}^{n}(k+1)*(k+1)!\]

Let j=k+1 so k=1, j=2, and k=n, j=n+1
Substitute k by j in the above
=\[\sum_{j=2}^{n+1}(j)*(j)!\]
=\[\sum_{j=1}^{n+1}(j)*(j)!\] - 1*1!
= (n+2)! - 1 - 1

So G(n) + 2*(n+1)! - 2 = (n+2)! - 2
G(n) = (n+2)! - 2*(n+1)!
= (n+1)!*(n + 2 - 2)
= n(n+1)!

which matches w what u found on wolfram @rational

It was such a stupid mistake that I made!!!! :(

Answer 31

that looks neat!! thank you