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Difference between revisions of "1969 AHSME Problems/Problem 16"

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== Solution ==
 
== Solution ==
 
Since <math>a=kb</math>, we can write <math>(a-b)^n</math> as <math>(kb-b)^n</math>.
 
Since <math>a=kb</math>, we can write <math>(a-b)^n</math> as <math>(kb-b)^n</math>.
−
Expanding, the second term is <math>-k^{n-1}b^{n}{n}\choose{1}</math>, and the third term is <math>k^{n-2}b^{n}{n}\choose{2}</math>, so we can write the equation
+
Expanding, the second term is <math>-k^{n-1}b^{n}{{n}\choose{1}}</math>, and the third term is <math>k^{n-2}b^{n}{{n}\choose{2}}</math>, so we can write the equation
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<math>-k^{n-1}b^{n}{n}\choose{1}+k^{n-2}b^{n}{n}\choose{2}=0</math>
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<cmath>-k^{n-1}b^{n}{{n}\choose{1}}+k^{n-2}b^{n}{{n}\choose{2}}=0</cmath>
 
Simplifying and multiplying by two to remove the denominator, we get
 
Simplifying and multiplying by two to remove the denominator, we get
−
<math>-2k^{n-1}b^{n}n+k^{n-2}b^{n}n(n+1)=0</math>
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<cmath>-2k^{n-1}b^{n}n+k^{n-2}b^{n}n(n-1)=0</cmath>
 
Factoring, we get
 
Factoring, we get
−
<math>k^{n-2}b^{n}n(-2k+n+1)=0</math>
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<cmath>k^{n-2}b^{n}n(-2k+n-1)=0</cmath>
−
Dividing by <math>k^{n-2}b^{n}n</math> gives
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Dividing by <math>k^{n-2}b^{n}</math> gives
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<math>n(-2k+n+1)=0</math>.
+
<cmath>n(-2k+n-1)=0</cmath>
 
Since it is given that <math>n\ge2</math>, <math>n</math> cannot equal 0, so we can divide by n, which gives
 
Since it is given that <math>n\ge2</math>, <math>n</math> cannot equal 0, so we can divide by n, which gives
−
<math>-2k+n+1=0.
+
<cmath>-2k+n-1=0</cmath>
−
Solving for </math>n<math> gives
+
Solving for <math>n</math> gives
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</math>n=2k-1<math>, so the answer is </math>\fbox{C}$.
+
<cmath>n=2k+1</cmath> so the answer is <math>\fbox{E}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 12:16, 11 October 2015

Problem

When $(a-b)^n,n\ge2,ab\ne0$, is expanded by the binomial theorem, it is found that when $a=kb$, where $k$ is a positive integer, the sum of the second and third terms is zero. Then $n$ equals:

$\text{(A) } \tfrac{1}{2}k(k-1)\quad \text{(B) } \tfrac{1}{2}k(k+1)\quad \text{(C) } 2k-1\quad \text{(D) } 2k\quad \text{(E) } 2k+1$

Solution

Since $a=kb$, we can write $(a-b)^n$ as $(kb-b)^n$. Expanding, the second term is $-k^{n-1}b^{n}{{n}\choose{1}}$, and the third term is $k^{n-2}b^{n}{{n}\choose{2}}$, so we can write the equation \[-k^{n-1}b^{n}{{n}\choose{1}}+k^{n-2}b^{n}{{n}\choose{2}}=0\] Simplifying and multiplying by two to remove the denominator, we get \[-2k^{n-1}b^{n}n+k^{n-2}b^{n}n(n-1)=0\] Factoring, we get \[k^{n-2}b^{n}n(-2k+n-1)=0\] Dividing by $k^{n-2}b^{n}$ gives \[n(-2k+n-1)=0\] Since it is given that $n\ge2$, $n$ cannot equal 0, so we can divide by n, which gives \[-2k+n-1=0\] Solving for $n$ gives \[n=2k+1\] so the answer is $\fbox{E}$.

See also

1969 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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