# Pentagonal Numbers Formula

• Apr 29th 2007, 03:17 AM
disco
Pentagonal Numbers Formula
Hi everyone

I have been asked to find General term for Pentagonal numbers in one of our Exams. I have solved the general term for the pentagonal numbers by subtracting the same series from itself. The procedure is as follows

S = 1 + 5 + 12 + 22 + 35 + .......................................+ tn
S = 0 + 1 + 5 + 12 + 22 + 35 + .......................................+ tn
( _ )
----------------------------------------------------------------------------------------------------
0 = 1 + 4 + 7 + 10 + 13 + ............................................. - tn
=> tn = 1 + 4 + 7 + 10 + 13 + ............................................. [ Sum of n terms of A.P with common difference 3 and first term 1]
tn = (n/2) (3n - 1)

I hope the method is absolutely correct to find the general term....
Am I correct in doing so???

If I am correct Then ----
I have got 0 [zero] marks out of 6 marks for this question
My Instructor is not convinced with the method that i have adopted i.e subtracting a series from the same series itself ....
He says it can be implemented only for finite series..

OR
Any link That supports my method ...

So that i can show the link to my instructor to get my marks...

Is there any specific name to the method that i have adopted ???
Examples which use this method are required urgently...???

I kindly request everyone to Give links that support my method

bye
• Apr 29th 2007, 06:36 AM
ThePerfectHacker
The n-th pentagonal number is: n(3n-1)/2
• Apr 29th 2007, 07:03 AM
Soroban
Hello, disco!

Quote:

I have been asked to find general term for Pentagonal numbers.
I have solved the general term for the pentagonal numbers
by subtracting the same series from itself.
The procedure is as follows:

S . = . 1 + 5 + 12 + 22 + 35 + . . . + t
n
S . = . . . . 1 .+ 5 + 12 + 22 + 35 + . . . + t
n

0 . = . 1 + 4 + 7 + 10 + 13 + . . . - t
n

t
n . = . 1 + 4 + 7 + 10 + 13 + . . .
. . [Sum of n terms of A.P with common difference 3 and first term 1]

t
n . = . n(3n - 1)/2
I see nothing wrong with this derivation!

Quote:

I have got 0 [zero] marks out of 6 marks for this question.
My Instructor is not convinced with the method that i have adopted.

Does he think you got the correct answer by sheer luck?

Quote:

He says it can be implemented only for finite series.

Your series is finite . . . it went to t
n.

You might expect to get nowhere with your instructor.
If he (or she) is the type who simply cannot stand "a different approach",
. . all the documentation in the world won't convince him.
Instead of complimenting you for an original derivation,
. . you are penalized for not using his method.

Over my career, I was constantly surprised by innovative solutions from my students.
At first glance, it looks like nonsense. .Then I see that the answer is correct.
. . Hmm, could it be correct after all?
Then I find that it is a valid solution, using a method I didn't teach.
. . Of course, he gets full credit!

Imagine . . . He forgot the two methods I demonstrated in class
. . and with the pressure of an exam, he created an innovative approach.
This remarkable solution goes into my personal math file.

The next time I teach this topic, I show my class the three methods,
. . pubicly thanking that student for his contribution.

• Apr 30th 2007, 04:56 AM
disco
HI Soroban

Sorry for the mistake that
He says it can be implemented only for finite series.

He says it can be implemented only for infinite series

OR
Any link That supports my method ...

I am badly in need of those 6 marks ...
I may miss my A grade because of that.........

Bye
• Apr 30th 2007, 06:07 AM
Soroban
Hello,disco!

Quote:

He says it can be implemented only for infinite series.
This is not true . . .

Suppose we did not know the formula for a triangular number.

We can create a sum of consecutive triangular numbers.

. . . . . . . .S .= .1 + 3 + 6 + 10 + 15 + . . . + t
n

Then: . . . S .= . . . .1 + 3 + .6 + 10 + 15 + . . . + t
n

Subtract: .0 .= .1 + 2 + 3 + 4 + 5 + . . . + n .- .t
n

Hence: .t
n .= .1 + 2 + 3 + 4 + . . . + n

And from here, we can derive: .t
n .= .n(n + 1)/2