Difference between revisions of "2002 AMC 10B Problems/Problem 23"
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== Solution 1 == | == Solution 1 == | ||
− | + | When <math>m=1</math>, <math>a_{n+1}=1+a_n+n</math>. Hence, | |
− | <cmath> | + | <cmath>a_{2}=1+a_1+2</cmath> |
− | + | <cmath>a_{3}=1+a_2+3</cmath> | |
− | + | <cmath>a_{4}=1+a_3+4</cmath> | |
− | + | <math>\dots</math> | |
− | + | <cmath>a_{12}=1+a_{11}+11</cmath> | |
− | <cmath> | + | Adding these equations up, we have that <math>a_12=12+(1+2+3+...+11)=\boxed{78}</math> |
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− | <cmath> | ||
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== Solution 2 == | == Solution 2 == |
Revision as of 22:31, 18 July 2019
Problem 23
Let be a sequence of integers such that
and
for all positive integers
and
Then
is
Solution 1
When ,
. Hence,
Adding these equations up, we have that
Solution 2
Substituting into
:
. Since
,
. Therefore,
, and so on until
. Adding the Left Hand Sides of all of these equations gives
; adding the Right Hand Sides of these equations gives
. These two expressions must be equal; hence
and
. Substituting
:
. Thus we have a general formula for
and substituting
:
.
Solution 3
We can literally just plug stuff in. No prerequisite is actually said in the sequence. Since , we know
. After this, we can use
to find
.
. Now, we can use
and
to find
, or
. Lastly, we can use
to find
.
Additional Comment
This is also the formula for the triangular numbers , as seen in Solution 2
See also
2002 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.