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2007 iTest Problems/Problem 58: Difference between revisions

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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..."
 
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Solution to Problem 58 — fun problem!
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== Problem ==
''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.''
Let <math>T=\text{TNFTPP}</math>. For natural numbers <math>k,n\geq 2</math>, we define
 
==Problem==
 
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>
Compute the value of <math>S(10,T+55)-S(10,55)+S(10,T-55)</math>.
Compute the value of <math>S(10,112)-S(10,55)+S(10,2)</math>.
 
==Solution==
 
The function <math>S(k,n)</math> can be rewritten as
<cmath>\sum_{j=2}^{k} \left\lfloor\frac{j^{n+1}+1}{j^{n-1}+1}\right\rfloor</cmath>
 
Let <math>x = j^{n-1}</math>.  With the substitution, each similar part becomes
<cmath>\sum_{j=2}^{k} \left\lfloor\frac{j^2 \cdot x+1}{x+1}\right\rfloor</cmath>
Performing polynomial division results in
<cmath>\sum_{j=2}^{k} \left\lfloor j^2 + \frac{1 - j^2}{x+1}\right\rfloor</cmath>
<cmath>\sum_{j=2}^{k} \left\lfloor j^2 - \frac{j^2 - 1}{j^{n-1}+1}\right\rfloor</cmath>
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>.
 
<br>
That means <math>S(10,112) - S(10,55) + S(10,2) = S(10,2)</math>, and that equals
<cmath>\sum_{j=2}^{10} \left\lfloor j^2 - \frac{j^2 - 1}{j+1}\right\rfloor</cmath>
<cmath>\sum_{j=2}^{10} \left\lfloor j^2 - (j-1) \right\rfloor</cmath>
<cmath>\sum_{j=2}^{10} j^2 - \sum_{j=2}^{10} j-1</cmath>
<cmath>(4+9 \cdots 100) - (1+2 \cdots 9)</cmath>
<cmath>(\frac{10 \cdot 11 \cdot 21}{6} - 1) - (\frac{10 \cdot 9}{2}</cmath>
<cmath>384-45</cmath>
<cmath>\boxed{339}</cmath>
 
==See Also==
{{iTest box|year=2007|num-b=57|num-a=59}}


== Solution ==
[[Category:Intermediate Algebra Problems]]

Revision as of 12:03, 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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