Difference between revisions of "2002 AMC 10B Problems/Problem 23"
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We can set <math>n</math> equal to <math>m</math>, so we can say that | We can set <math>n</math> equal to <math>m</math>, so we can say that | ||
− | <cmath>a_{m + m} = a_m + a_m | + | <cmath>a_{m + m} = a_m + a_m + m*m</cmath> |
<cmath>a_{2m} = 2a_m + m^2</cmath> | <cmath>a_{2m} = 2a_m + m^2</cmath> | ||
Revision as of 17:04, 27 December 2019
Contents
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
~AopsUser101
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
.
Solution 4
We can set equal to
, so we can say that
We set , we get
.
We set m, we get
.
Solving for is easy, just direct substitution.
Substituting, we get
Thus, the answer is .
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 |
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