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Sequence and Series

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by raleigh » Tue May 19, 2009 2:33 pm
Ignore the _____. I used them to get the spacing to line up for the equations :(
First, figure out what S(n) sums up to. Write S(n) in ascending order, then again in descending order and add them.

__S(n) = 1 + 2 + 3 + ... + n
+ S(n) = n + (n-1) + (n-2) + ... + 1

2S(n) = (n+1) + (n+1) + ... + (n + 1)

Note that you get n+1, n times.

So 2S(n) = n(n+1). Then S(n) = n(n+1)/2. Plug in 2n for n and we find that S(2n) = 2n(2n+1)/2

S(2n) - S(n) = 2n(2n+1)/2 - n(n+1)/2
__________= (4n^2 + 2n - n^2 - n)/2
__________= (3n^2 + n)/2
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by scoobydooby » Wed May 20, 2009 4:03 am
am getting n(n+1)/2 =(n^2+n)/2

S(2n)= (2+4+6+....+2n)

=> 2(1+2+3+...+n)
=>2*S(n)

S(2n) - S(n)
=>2*S(n)- S(n)=S(n)=n(n+1)/2
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by El Cucu » Wed May 20, 2009 6:10 am
scoobydooby wrote:am getting n(n+1)/2 =(n^2+n)/2
Hi Scoobydooby, could you expalin how you arrive to:

n(n+1)/2 =(n^2+n)/2

tks!
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by El Cucu » Wed May 20, 2009 6:15 am
"Note that you get n+1, n times. So 2S(n) = n(n+1). "

Can you explain how you arrive to this So 2S(n) = n(n+1).


[/quote]
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by scoobydooby » Wed May 20, 2009 6:30 am
El Cucu wrote:
scoobydooby wrote:am getting n(n+1)/2 =(n^2+n)/2
Hi Scoobydooby, could you expalin how you arrive to:

n(n+1)/2 =(n^2+n)/2

tks!
just expanded n*(n+1) to n^2+n, so n(n+1)/2 =(n^2+n)/2
but, obviously i am missing something, since what i get is not the OA
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by raleigh » Wed May 20, 2009 6:54 am
El Cucu wrote:"Note that you get n+1, n times. So 2S(n) = n(n+1). "

Can you explain how you arrive to this So 2S(n) = n(n+1).

[/quote]
Treat _ as a space. The way the terms line up makes it easier to see. This is a calculus trick for finding the value of the arithmetic sum.
__S(n) = 1 + _ 2 + _ 3 +_ ... _ + n
+ S(n) = n + (n-1) + (n-2) + ... + 1

2S(n) = (n+1) + (n+1) + ... + (n + 1)

Add the two equations. Look how I grouped the terms:
n+1 = n+1
(n-1) + 2 = n+1
(n-2) + 3 = n+1

You see that each group of terms sums to n+1. There are n terms in S(n) so you must have (n+1) n times. This is n(n+1) mathematically.
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by El Cucu » Wed May 20, 2009 9:59 am
You see that each group of terms sums to n+1. There are n terms in S(n) so you must have (n+1) n times. This is n(n+1) mathematically.[/quote]

Nice explanation Raleigh! tks!
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