# Thread: Find the 1000th term

1. ## Find the 1000th term

$1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5.............,k......$ is a sequence where the number $k$ appears $k$ times successively. $(k=1, 2, 3, 4, 5,...)$ Find the $1000th$ term of the sequence.

$S_k=1000$

$1+2+3+4+5+...+n=1000$

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

$\frac{n}{2}[n+1]=1000$

$n(n+1)=2000$

$n^2+n-2000=0$

$n=-45.22(reject) or 44.22$

should i accept n=44 or n=45 as the answer?

2. Originally Posted by Punch
$1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5.............,k......$ is a sequence where the number $k$ appears $k$ times successively. $(k=1, 2, 3, 4, 5,...)$ Find the $1000th$ term of the sequence.

$S_k=1000$

$1+2+3+4+5+...+n=1000$

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

$\frac{n}{2}[n+1]=1000$

$n(n+1)=2000$

$n^2+n-2000=0$

$n=-45.22(reject) or 44.22$

should i accept n=44 or n=45 as the answer?

I don't understood what you done, bur here is my hint:

Observe when 1,2,3,4,5,... appears in your sequence.

Try to find a close formula to where k first time appears

3. Originally Posted by Also sprach Zarathustra
I don't understood what you done, bur here is my hint:

Observe when 1,2,3,4,5,... appears in your sequence.

Try to find a close formula to where k first time appears
Notice that the question says a sequence where the number k appears k times successively.

So the sequence k=1,2,3,4,5,...... and so on.

Sequence k is an arithmetic progression.

Since the number k appears k times, the sum of the sequence up to the kth term is 1000.

Hence, 1+2+3+4+5+...+n=1000

then by using the sum of an arithmetic progression, i used it to find n

4. Here is close formula for your sequence:

$f(n)=\left \lfloor {\frac{1+\sqrt{1+8n}}{2}} \right \rfloor$

5. Originally Posted by Punch
$1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5.............,k......$ is a sequence where the number $k$ appears $k$ times successively. $(k=1, 2, 3, 4, 5,...)$ Find the $1000th$ term of the sequence.

$S_k=1000$

$1+2+3+4+5+...+n=1000$

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

$\frac{n}{2}[n+1]=1000$

$n(n+1)=2000$

$n^2+n-2000=0$

$n=-45.22(reject) or 44.22$

should i accept n=44 or n=45 as the answer?
Suppose the 1000th term is 44 count the maximum number of terms less than or equal to 44...

The largest number of terms less than or equal to 44 is:

1+2+ ... +44=990,

a contradiction, now try a similar argument with 45.

CB

6. Originally Posted by CaptainBlack
Suppose the 1000th term is 44 count the maximum number of terms less than or equal to 44...

The largest number of terms less than or equal to 44 is:

1+2+ ... +44=990,

a contradiction, now try a similar argument with 45.

CB
Oh yes, i could have used the sum formula

7. Originally Posted by Punch
$1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5.............,k......$ is a sequence where the number $k$ appears $k$ times successively. $(k=1, 2, 3, 4, 5,...)$ Find the $1000th$ term of the sequence.

$S_k=1000$

$1+2+3+4+5+...+n=1000$
What does this have to do with your problem?

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

$\frac{n}{2}[n+1]=1000$

$n(n+1)=2000$

$n^2+n-2000=0$

$n=-45.22(reject) or 44.22$

should i accept n=44 or n=45 as the answer?

8. Originally Posted by Punch
$1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5.............,k......$ is a sequence where the number $k$ appears $k$ times successively. $(k=1, 2, 3, 4, 5,...)$ Find the $1000th$ term of the sequence.

$S_k=1000$

$1+2+3+4+5+...+n=1000$

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

$\frac{n}{2}[n+1]=1000$

$n(n+1)=2000$

$n^2+n-2000=0$

$n=-45.22(reject) or 44.22$

should i accept n=44 or n=45 as the answer?
You are taking the last of the repeated numbers and calling it Tn.

T1=1
T3=2, and 1+2=3
T6=3, and 1+2+3=6
T10=4, and 1+2+3+4=10

Hence you want T1000=1+2+3+4+....+n.

However, you do not know that T1000 is the last of a repeated list of the same number.

T990=44
generated from the sum of terms, so it's the last 44.
Hence T991 is 45 and how many of those follow T990 ?

9. Originally Posted by Archie Meade
You are taking the last of the repeated numbers and calling it Tn.

T1=1
T3=2, and 1+2=3
T6=3, and 1+2+3=6
T10=4, and 1+2+3+4=10

Hence you want T1000=1+2+3+4+....+n.

However, you do not know that T1000 is the last of a repeated list of the same number.