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Difference between revisions of "2007 iTest Problems/Problem 58"

(Created page with "== Problem == Let <math>T=\text{TNFTPP}</math>. For natural numbers <math>k,n\geq 2</math>, we define <cmath>S(k,n)=\left\lfloor\frac{2^{n+1}+1}{2^{n-1}+1}\right\rfloor+\left\lfl...")
 
(Solution to Problem 58 — fun problem!)
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== Problem ==
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''The following problem is from the Ultimate Question of the [[2007 iTest]], where solving this problem required the answer of a previous problem.  When the problem is rewritten, the T-value is substituted.''
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Let <math>T=\text{TNFTPP}</math>. For natural numbers <math>k,n\geq 2</math>, we define
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==Problem==
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For natural numbers <math>k,n\geq 2</math>, we define
 
<cmath>S(k,n)=\left\lfloor\frac{2^{n+1}+1}{2^{n-1}+1}\right\rfloor+\left\lfloor\frac{3^{n+1}+1}{3^{n-1}+1}\right\rfloor+\cdots+\left\lfloor\frac{k^{n+1}+1}{k^{n-1}+1}\right\rfloor</cmath>
 
<cmath>S(k,n)=\left\lfloor\frac{2^{n+1}+1}{2^{n-1}+1}\right\rfloor+\left\lfloor\frac{3^{n+1}+1}{3^{n-1}+1}\right\rfloor+\cdots+\left\lfloor\frac{k^{n+1}+1}{k^{n-1}+1}\right\rfloor</cmath>
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Compute the value of <math>S(10,T+55)-S(10,55)+S(10,T-55)</math>.
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Compute the value of <math>S(10,112)-S(10,55)+S(10,2)</math>.
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==Solution==
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The function <math>S(k,n)</math> can be rewritten as
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<cmath>\sum_{j=2}^{k} \left\lfloor\frac{j^{n+1}+1}{j^{n-1}+1}\right\rfloor</cmath>
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Let <math>x = j^{n-1}</math>.  With the substitution, each similar part becomes
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<cmath>\sum_{j=2}^{k} \left\lfloor\frac{j^2 \cdot x+1}{x+1}\right\rfloor</cmath>
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Performing polynomial division results in
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<cmath>\sum_{j=2}^{k} \left\lfloor j^2 + \frac{1 - j^2}{x+1}\right\rfloor</cmath>
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<cmath>\sum_{j=2}^{k} \left\lfloor j^2 - \frac{j^2 - 1}{j^{n-1}+1}\right\rfloor</cmath>
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When <math>n = 112</math> or <math>n = 55</math>, then <math>\frac{j^2 - 1}{j^{n-1}+1}</math> is close to zero, which means that <math>\left\lfloor j^2 - \frac{j^2 - 1}{j^{n-1}+1}\right\rfloor</math> would be the same for a given <math>j</math> when <math>n = 112</math> or <math>n = 55</math>.  Thus, <math>S(10,112) - S(10,55) = 0</math>.
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<br>
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That means <math>S(10,112) - S(10,55) + S(10,2) = S(10,2)</math>, and that equals
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<cmath>\sum_{j=2}^{10} \left\lfloor j^2 - \frac{j^2 - 1}{j+1}\right\rfloor</cmath>
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<cmath>\sum_{j=2}^{10} \left\lfloor j^2 - (j-1) \right\rfloor</cmath>
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<cmath>\sum_{j=2}^{10} j^2 - \sum_{j=2}^{10} j-1</cmath>
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<cmath>(4+9 \cdots 100) - (1+2 \cdots 9)</cmath>
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<cmath>(\frac{10 \cdot 11 \cdot 21}{6} - 1) - (\frac{10 \cdot 9}{2}</cmath>
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<cmath>384-45</cmath>
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<cmath>\boxed{339}</cmath>
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==See Also==
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{{iTest box|year=2007|num-b=57|num-a=59}}
  
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== Solution ==
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[[Category:Intermediate Algebra Problems]]

Revision as of 13:03, 30 June 2018

The following problem is from the Ultimate Question of the 2007 iTest, where solving this problem required the answer of a previous problem. When the problem is rewritten, the T-value is substituted.

Problem

For natural numbers $k,n\geq 2$, we define \[S(k,n)=\left\lfloor\frac{2^{n+1}+1}{2^{n-1}+1}\right\rfloor+\left\lfloor\frac{3^{n+1}+1}{3^{n-1}+1}\right\rfloor+\cdots+\left\lfloor\frac{k^{n+1}+1}{k^{n-1}+1}\right\rfloor\] Compute the value of $S(10,112)-S(10,55)+S(10,2)$.

Solution

The function $S(k,n)$ can be rewritten as \[\sum_{j=2}^{k} \left\lfloor\frac{j^{n+1}+1}{j^{n-1}+1}\right\rfloor\]

Let $x = j^{n-1}$. With the substitution, each similar part becomes \[\sum_{j=2}^{k} \left\lfloor\frac{j^2 \cdot x+1}{x+1}\right\rfloor\] Performing polynomial division results in \[\sum_{j=2}^{k} \left\lfloor j^2 + \frac{1 - j^2}{x+1}\right\rfloor\] \[\sum_{j=2}^{k} \left\lfloor j^2 - \frac{j^2 - 1}{j^{n-1}+1}\right\rfloor\] When $n = 112$ or $n = 55$, then $\frac{j^2 - 1}{j^{n-1}+1}$ is close to zero, which means that $\left\lfloor j^2 - \frac{j^2 - 1}{j^{n-1}+1}\right\rfloor$ would be the same for a given $j$ when $n = 112$ or $n = 55$. Thus, $S(10,112) - S(10,55) = 0$.


That means $S(10,112) - S(10,55) + S(10,2) = S(10,2)$, and that equals \[\sum_{j=2}^{10} \left\lfloor j^2 - \frac{j^2 - 1}{j+1}\right\rfloor\] \[\sum_{j=2}^{10} \left\lfloor j^2 - (j-1) \right\rfloor\] \[\sum_{j=2}^{10} j^2 - \sum_{j=2}^{10} j-1\] \[(4+9 \cdots 100) - (1+2 \cdots 9)\] \[(\frac{10 \cdot 11 \cdot 21}{6} - 1) - (\frac{10 \cdot 9}{2}\] \[384-45\] \[\boxed{339}\]

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 57 		Followed by:
Problem 59
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