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

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(A) <math>-2^{50}</math> (B) <math>-2^{49}</math> (C) 0 (D) <math>2^{49}</math> (E) <math>2^{50}</math>
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<math> \textrm{(A)}\ -2^{50}\qquad\textrm{(B)}\ -2^{49}\qquad\textrm{(C)}\ 0\qquad\textrm{(D)}\ 2^{49}\qquad\textrm{(E)}\ 2^{50} </math>
  
 
==Solution==
 
==Solution==

Latest revision as of 12:32, 29 July 2025

Problem

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


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

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, $(1+i)^{99}=[\sqrt{2}\mathrm{cis}(45^\circ)]^{99}=\sqrt{2^{99}}\cdot \mathrm{cis}(99\cdot45^\circ)=2^{49}\sqrt{2}\cdot \mathrm{cis}(135^\circ) = -2^{49}+2^{49}i$.

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

$\fbox{B}$

See also

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

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