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- #1

- Feb 14, 2012

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What are the last four digits of $a_{2012}$?

- Thread starter anemone
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- Thread starter
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- #1

- Feb 14, 2012

- 3,682

What are the last four digits of $a_{2012}$?

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- #2

- Feb 7, 2012

- 2,703

What are the last four digits of $a_{2012}$?

- Aug 30, 2012

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@Opalg:

-Dan

-Dan

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- #4

- Mar 5, 2012

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Nice!

But I'll pick 2037 as answer.

But I'll pick 2037 as answer.

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- #5

- Feb 7, 2012

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Erm, yes.Nice!

But I'll pick 2037 as answer.

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- #6

- Feb 14, 2012

- 3,682

Thanks for participating **Opalg**!

Your method seems so perfect to me.

My solution:

Observe that

$a_1=1$, $a_2=2$, $a_3=3$, $a_4=5$,$a_5=7$, $a_6=10$, $a_7=13$, $a_8=17$, $a_9=21$, $a_{10}=26$, $\cdots$

We are asked to find the last 4 digits of $a_{2012}$ term, so, if we only consider the terms $a_2=2$, $a_4=5$, $a_6=10$, $a_8=17$, $a_{10}=26$, $\cdots$, we see that

1.

the differences between the terms made up an arithemetic sequence (3, 5, 7, 9, ......), since

and hence we get the sum of the first $n$ terms of the sequence (3, 5, 7, 9,......) be

$S_n=\dfrac{n}{2}(2(3)+(n-1)(2))=n(n+2)$.

2. Once we recognize this pattern, we can easily find whichever term that we would like to, says, if we want to look for the 5th term in this sequence ($2,5,10,17,26,37,\cdots$), then what are we going to do to find that fifth term is by adding the first term, that is 2 to the sum of the first 4 terms of the arithmetic sequence (3, 5, 7, 9, ......), i.e.

$a_5=2+4(4+2)=26$

Therefore, if we look for the 1006th term in the sequence ($a_2=2$, $a_4=5$, $a_6=10$, $a_8=17$, $a_{10}=26$, $\cdots$), that means we are actually find the $a_{2012}$ of the given sequence in the original problem.

So, $a_{1006}=2+1005(1005+2)=1012037$, thus, the last four digits of $a_{2012}$ is 2037.

Your method seems so perfect to me.

My solution:

Observe that

$a_1=1$, $a_2=2$, $a_3=3$, $a_4=5$,$a_5=7$, $a_6=10$, $a_7=13$, $a_8=17$, $a_9=21$, $a_{10}=26$, $\cdots$

We are asked to find the last 4 digits of $a_{2012}$ term, so, if we only consider the terms $a_2=2$, $a_4=5$, $a_6=10$, $a_8=17$, $a_{10}=26$, $\cdots$, we see that

1.

the differences between the terms made up an arithemetic sequence (3, 5, 7, 9, ......), since

and hence we get the sum of the first $n$ terms of the sequence (3, 5, 7, 9,......) be

$S_n=\dfrac{n}{2}(2(3)+(n-1)(2))=n(n+2)$.

2. Once we recognize this pattern, we can easily find whichever term that we would like to, says, if we want to look for the 5th term in this sequence ($2,5,10,17,26,37,\cdots$), then what are we going to do to find that fifth term is by adding the first term, that is 2 to the sum of the first 4 terms of the arithmetic sequence (3, 5, 7, 9, ......), i.e.

$a_5=2+4(4+2)=26$

Therefore, if we look for the 1006th term in the sequence ($a_2=2$, $a_4=5$, $a_6=10$, $a_8=17$, $a_{10}=26$, $\cdots$), that means we are actually find the $a_{2012}$ of the given sequence in the original problem.

So, $a_{1006}=2+1005(1005+2)=1012037$, thus, the last four digits of $a_{2012}$ is 2037.

Last edited:

- Mar 31, 2013

- 1,309

only one word is sufficient to describe your solutionThanks for participatingOpalg!

Your method seems so perfect to me.

My solution:

Observe that

$a_1=1$, $a_2=2$, $a_3=3$, $a_4=5$,$a_5=7$, $a_6=10$, $a_7=13$, $a_8=17$, $a_9=21$, $a_{10}=26$, $\cdots$

We are asked to find the last 4 digits of $a_{2012}$ term, so, if we only consider the terms $a_2=2$, $a_4=5$, $a_6=10$, $a_8=17$, $a_{10}=26$, $\cdots$, we see that

1.

the differences between the terms made up an arithemetic sequence (3, 5, 7, 9, ......), since

View attachment 1659

and hence we get the sum of the first $n$ terms of the sequence (3, 5, 7, 9,......) be

$S_n=\dfrac{n}{2}(2(3)+(n-1)(2))=n(n+2)$.

2. Once we recognize this pattern, we can easily find whichever term that we would like to, says, if we want to look for the 5th term in this sequence ($2,5,10,17,26,37,\cdots$), then what are we going to do to find that fifth term is by adding the first term, that is 2 to the sum of the first 4 terms of the arithmetic sequence (3, 5, 7, 9, ......), i.e.

$a_5=2+4(4+2)=26$

Therefore, if we look for the 1006th term in the sequence ($a_2=2$, $a_4=5$, $a_6=10$, $a_8=17$, $a_{10}=26$, $\cdots$), that means we are actually find the $a_{2012}$ of the given sequence in the original problem.

So, $a_{1006}=2+1006(1006+2)=1012037$, thus, the last four digits of $a_{2012}$ is 2037.

elegant

- Jan 25, 2013

- 1,225

it should be :Thanks for participatingOpalg!

Your method seems so perfect to me.

My solution:

Observe that

$a_1=1$, $a_2=2$, $a_3=3$, $a_4=5$,$a_5=7$, $a_6=10$, $a_7=13$, $a_8=17$, $a_9=21$, $a_{10}=26$, $\cdots$

We are asked to find the last 4 digits of $a_{2012}$ term, so, if we only consider the terms $a_2=2$, $a_4=5$, $a_6=10$, $a_8=17$, $a_{10}=26$, $\cdots$, we see that

1.

the differences between the terms made up an arithemetic sequence (3, 5, 7, 9, ......), since

View attachment 1659

and hence we get the sum of the first $n$ terms of the sequence (3, 5, 7, 9,......) be

$S_n=\dfrac{n}{2}(2(3)+(n-1)(2))=n(n+2)$.

2. Once we recognize this pattern, we can easily find whichever term that we would like to, says, if we want to look for the 5th term in this sequence ($2,5,10,17,26,37,\cdots$), then what are we going to do to find that fifth term is by adding the first term, that is 2 to the sum of the first 4 terms of the arithmetic sequence (3, 5, 7, 9, ......), i.e.

$a_5=2+4(4+2)=26$

Therefore, if we look for the 1006th term in the sequence ($a_2=2$, $a_4=5$, $a_6=10$, $a_8=17$, $a_{10}=26$, $\cdots$), that means we are actually find the $a_{2012}$ of the given sequence in the original problem.

So, $a_{1006}=2+1006(1006+2)=1012037$, thus, the last four digits of $a_{2012}$ is 2037.

$a_{1006}=2+1005(1005+2)=1012037$, thus, the last four digits of $a_{2012}$ is 2037

- Thread starter
- Admin
- #9

- Feb 14, 2012

- 3,682

Oops! You're absolutely right! I will fix my previous post and thank you so much for pointing this out,it should be :

$a_{1006}=2+1005(1005+2)=1012037$, thus, the last four digits of $a_{2012}$ is 2037