Difference between revisions of "1989 AHSME Problems/Problem 29"

Revision as of 13:25, 7 October 2014

Problem

What is the value of the sum $S=\sum_{k=0}^{49}(-1)^k\binom{99}{2k}=\binom{99}{0}-\binom{99}{2}+\binom{99}{4}-\cdots -\binom{99}{98}?$

(A) $-2^{50}$ (B) $-2^{49}$ (C) 0 (D) $2^{49}$ (E) $2^{50}$

Solution

By the Binomial Theorem, $(1+i)^{99}=\sum_{n=0}^{99}\binom{99}{j}i^n =$ $\binom{99}{0}i^0+\binom{99}{1}i^1+\binom{99}{2}i^2+\binom{99}{3}i^3+\binom{99}{4}i^4+\cdots +\binom{99}{98}i^{98}$ .

Using the fact that $i^1=i$ , $i^2=-1$ , $i^3=-i$ , $i^4=1$ , and $i^{n+4}=i^n$ , the sum becomes:

$(1+i)^{99}=\binom{99}{0}+\binom{99}{1}i-\binom{99}{2}-\binom{99}{3}i+\binom{99}{4}+\cdots -\binom{99}{98}$ .

So, $Re[(1+i)^{99}]=\binom{99}{0}-\binom{99}{2}+\binom{99}{4}-\cdots -\binom{99}{98} = S$ .

Using De Moivre's Theorem, $(1+i)^{99}=[\sqrt{2}cis(45^\circ)]^{99}=\sqrt{2^{99}}\cdot cis(99\cdot45^\circ)=2^{49}\sqrt{2}\cdot cis(135^\circ) = -2^{49}+2^{49}i$ .

And finally, $S=Re[-2^{49}+2^{49}i] = -2^{49}$ .

$\fbox{B}$

1989 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 28	Followed by Problem 30
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 16: / Line 16: @@
 Using [[De Moivre's Theorem]], <math>(1+i)^{99}=[\sqrt{2}cis(45^\circ)]^{99}=\sqrt{2^{99}}\cdot cis(99\cdot45^\circ)=2^{49}\sqrt{2}\cdot cis(135^\circ) = -2^{49}+2^{49}i</math>.
 And finally, <math>S=Re[-2^{49}+2^{49}i] = -2^{49}</math>.
+<math>\fbox{B}</math>
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Difference between revisions of "1989 AHSME Problems/Problem 29"

Revision as of 13:25, 7 October 2014

Problem

Solution

See also