Difference between revisions of "2009 AMC 12A Problems/Problem 15"
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== Solution 1 == | == Solution 1 == | ||
− | We know that <math>i^x</math> cycles every <math>4</math> | + | We know that <math>i^x</math> cycles every <math>4</math> powers so we group the sum in <math>4</math>s. |
<cmath>i+2i^2+3i^3+4i^4=2-2i</cmath> | <cmath>i+2i^2+3i^3+4i^4=2-2i</cmath> | ||
<cmath>5i^5+6i^6+7i^7+8i^8=2-2i</cmath> | <cmath>5i^5+6i^6+7i^7+8i^8=2-2i</cmath> | ||
Line 14: | Line 14: | ||
For 24 groups we thus, get <math>48-48i</math> as our sum. | For 24 groups we thus, get <math>48-48i</math> as our sum. | ||
We know the solution must lie near | We know the solution must lie near | ||
− | The next term is the <math>24*4+1=97</math>th term. This term is equal to <math>97i</math> (first in a group of <math>4</math>) and our sum is now <math>48+49i</math> so <math>n=97</math> is our answer | + | The next term is the <math>24*4+1=97</math>th term. This term is equal to <math>97i</math> (first in a group of <math>4</math> so <math>i^{97}=i</math>) and our sum is now <math>48+49i</math> so <math>n=97</math> is our answer |
− | |||
== Solution 2== | == Solution 2== |
Revision as of 07:53, 8 March 2019
Contents
Problem
For what value of is
?
Note: here .
Solution 1
We know that cycles every
powers so we group the sum in
s.
We can postulate that every group of is equal to
.
For 24 groups we thus, get
as our sum.
We know the solution must lie near
The next term is the
th term. This term is equal to
(first in a group of
so
) and our sum is now
so
is our answer
Solution 2
Obviously, even powers of are real and odd powers of
are imaginary.
Hence the real part of the sum is
, and
the imaginary part is
.
Let's take a look at the real part first. We have , hence the real part simplifies to
.
If there were an odd number of terms, we could pair them as follows:
, hence the result would be negative. As we need the real part to be
, we must have an even number of terms. If we have an even number of terms, we can pair them as
. Each parenthesis is equal to
, thus there are
of them, and the last value used is
. This happens for
and
. As
is not present as an option, we may conclude that the answer is
.
In a complete solution, we should now verify which of and
will give us the correct imaginary part.
We can rewrite the imaginary part as follows: . We need to obtain
. Once again we can repeat the same reasoning: If the number of terms were even, the left hand side would be negative, thus the number of terms is odd. The left hand side can then be rewritten as
. We need
parentheses, therefore the last value used is
. This happens when
or
, and we are done.
See Also
2009 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.