Question

Let sequence $(a_{n}): 0,1,1,2,2,3,3,4,4,....,k,k,k+1,k+1,.....$.
Compute the sum $S_{n}=a_{1}+a_{2}+.....+a_{n}$.

Answer 1

\[n(n+1)\times\frac{1}{2}\times 2\]

Answer 2

isn't this just twice the sum of a simple arithmetic progression: 1, 2, 3, 4, ..., n?

Answer 3

if you were adding up from 1 to 2n

Answer 4

oh yes, so its twice the sum of: 1, 2, 3, ..., n/2

Answer 5

it is thrown off a little since the first value is 0

Answer 6

but the zero seems to be ignored because it says S = a(1) + a(2) + ...

Answer 7

is \[a_1\text{ or }a_0\] the first term?

Answer 8

\[a_1\]I see what you are saying now, so a1 IS zero

Answer 9

that's what I think

Answer 10

so my guess is the sum will involve some mod operators to check for odd/even n?

Answer 11

yes

Answer 12

ok, so taking @Zarkon's advice, here is my attempt at this:
if n is odd, then there are (n+1)/2 terms starting with 0 and (n-1)/2 terms starting with 1. so we get the Odd N Sum (ONS) as:\[ONS_n=\frac{n+1}{4}(0+(\frac{n+1}{2}-1))+\frac{n-1}{4}(2+(\frac{n-1}{2}-1))\]\[=\frac{n^2-1}{8}\]
if n is even, then there are n/2 terms starting with 0 and n/2 terms starting with 1. so we get the Even N Sum (ENS) as:\[ENS_n=\frac{n}{4}(0+(\frac{n}{2}-1))+\frac{n}{4}(2+(\frac{n}{2}-1))\]\[=\frac{n^2}{4}\]
so:
n MOD 2 == 0 ==> S_n = ESN_n = n^2/4
n MOD 2 != 0 ==> S_n = ONS_n = (n^2-1)/8

Answer 13

these all make use of the sum of an arithmetic progression to 'n' terms where the difference between each term is 'd' and the first term is 'a' being:\[S_n=\frac{n}{2}(2a+(n-1)d)\]in this case 'd' is 1 for both sequences and the first term 'a' is 0 for one sequence and 1 for the other.

Answer 14

sorry I wrote the sum for Odd 'n' incorrectly, it be (n^2-1)/4

I guess we could simplify it further by writing the sum as:

S_n = (n^2 - (n MOD 2))/4

which takes care of the odd/even nature of 'n'.

Answer 15

does anyone know of how to enter MOD in the equation editor?

Answer 16

I managed to work out how to enter mod in Latex, so the sum can be simplified to:\[S_n=\frac{n^2-(n\mod{2})}{4}\]

Answer 17

\[s_{n}=n^{2}/4 , or (n^{2}-1)/4\]

Answer 18

for even values of n, \(S_n=\frac{n^2}{4}\), and for odd values of n, \(S_n=\frac{n^2-1}{4}\). these can be combined into a single expression using the "mod" operator as I showed above.