Sequence and Series

[El Cucu]

Sn = 1+2+3+4+5...n
What is the value of S(2n) - S(n)?

Answer is [spoiler](n^2-n)/2[/spoiler]

Some math guru who can explain the process in detail, tks!

[raleigh]

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

[scoobydooby]

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

[El Cucu]

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!

[El Cucu]

"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]

[scoobydooby]

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

[raleigh]

"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.

[El Cucu]

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!