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Difference between revisions of "2024 AMC 12A Problems/Problem 21"

(Created page with "==Problem== Suppose that <math>a_1 = 2</math> and the sequence <math>(a_n)</math> satisfies the recurrence relation <cmath>\frac{a_n -1}{n-1}=\frac{a_{n-1}+1}{(n-1)+1}</cmath>...")
 
(solution)
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<math>\textbf{(A) } 338{,}550 \qquad \textbf{(B) } 338{,}551 \qquad \textbf{(C) } 338{,}552 \qquad \textbf{(D) } 338{,}553 \qquad \textbf{(E) } 338{,}554</math>
 
<math>\textbf{(A) } 338{,}550 \qquad \textbf{(B) } 338{,}551 \qquad \textbf{(C) } 338{,}552 \qquad \textbf{(D) } 338{,}553 \qquad \textbf{(E) } 338{,}554</math>
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 +
==Solution==
 +
Multiply both sides of the recurrence to find that <math>n(a_n-1)=(n-1)(a_{n-1}+1)=(n-1)(a_{n-1}-1)+(n-1)(2)</math>.
 +
 +
Let <math>b_n=n(a_n-1)</math>. Then the previous relation becomes
 +
<cmath>b_n=b_{n-1}+2(n-1)</cmath>
 +
 +
We can rewrite this relation for values of <math>n</math> until <math>1</math> and use telescoping to derive an explicit formula:
 +
 +
<cmath>b_n=b_{n-1}+2(n-1)</cmath>
 +
<cmath>b_{n-1}=b_{n-2}+2(n-2)</cmath>
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<cmath>b_{n-2}=b_{n-3}+2(n-3)</cmath>
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<cmath>\cdot</cmath>
 +
<cmath>\cdot</cmath>
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<cmath>\cdot</cmath>
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<cmath>b_2=b_1+2(1)</cmath>
 +
 +
Summing the equations yields:
 +
 +
<cmath>b_n+b_{n-1}+\cdots+b_2=b_{n-1}+b_{n-2}+\cdots+b_1+2((n-1)+(n-2)+\cdots+1)</cmath>
 +
<cmath>b_n-b_1=2\cdot\frac{n(n-1)}{2}</cmath>
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<cmath>b_n-1=n(n-1)</cmath>
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<cmath>b_n=n(n-1)+1</cmath>
 +
 +
Now we can substitute <math>a_n</math> back into our equation:
 +
 +
<cmath>n(a_n-1)=n(n-1)+1</cmath>
 +
<cmath>a_n-1=n-1+\frac{1}{n}</cmath>
 +
<cmath>a_n=n+\frac{1}{n}</cmath>
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<cmath>a_n^2=n^2+\frac{1}{n^2}+2</cmath>
 +
 +
Thus the sum becomes
 +
 +
<cmath>\sum^{100}_{n=1} a_n^2= \sum^{100}_{n=1} n^2+ \sum^{100}_{n=1} \frac{1}{n^2}+ \sum^{100}_{n=1} 2</cmath>
 +
 +
We know that <math>\sum^{100}_{n=1} n^2=\frac{100\cdot101\cdot201}{6}=338350</math>, and we also know that <math>\sum^{100}_{n=1} 2=200</math>, so the requested sum is equivalent to <math>\sum^{100}_{n=1} \frac{1}{n^2}+338550</math>. All that remains is to calculate <math>\sum^{100}_{n=1} \frac{1}{n^2}</math>, and we know that this value lies between <math>1</math> and <math>2</math> (see the note below for a proof). Thus,
 +
 +
<cmath>\sum^{100}_{n=1} a_n^2= \sum^{100}_{n=1} \frac{1}{n^2}+338550<2+338550=338552</cmath>
 +
<cmath>\sum^{100}_{n=1} a_n^2= \sum^{100}_{n=1} \frac{1}{n^2}+338550>1+338550=338551</cmath>
 +
 +
so
 +
 +
<cmath>\sum^{100}_{n=1} a_n^2\in(338551,338552)</cmath>
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 +
and thus the answer is <math>\boxed{\textbf{(B) }338551}</math>.
 +
 +
~eevee9406
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2024|ab=A|num-b=20|num-a=22}}
 
{{AMC12 box|year=2024|ab=A|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:21, 8 November 2024

Problem

Suppose that $a_1 = 2$ and the sequence $(a_n)$ satisfies the recurrence relation \[\frac{a_n -1}{n-1}=\frac{a_{n-1}+1}{(n-1)+1}\]for all $n \ge 2.$ What is the greatest integer less than or equal to \[\sum^{100}_{n=1} a_n^2?\]

$\textbf{(A) } 338{,}550 \qquad \textbf{(B) } 338{,}551 \qquad \textbf{(C) } 338{,}552 \qquad \textbf{(D) } 338{,}553 \qquad \textbf{(E) } 338{,}554$

Solution

Multiply both sides of the recurrence to find that $n(a_n-1)=(n-1)(a_{n-1}+1)=(n-1)(a_{n-1}-1)+(n-1)(2)$.

Let $b_n=n(a_n-1)$. Then the previous relation becomes \[b_n=b_{n-1}+2(n-1)\]

We can rewrite this relation for values of $n$ until $1$ and use telescoping to derive an explicit formula:

\[b_n=b_{n-1}+2(n-1)\] \[b_{n-1}=b_{n-2}+2(n-2)\] \[b_{n-2}=b_{n-3}+2(n-3)\] \[\cdot\] \[\cdot\] \[\cdot\] \[b_2=b_1+2(1)\]

Summing the equations yields:

\[b_n+b_{n-1}+\cdots+b_2=b_{n-1}+b_{n-2}+\cdots+b_1+2((n-1)+(n-2)+\cdots+1)\] \[b_n-b_1=2\cdot\frac{n(n-1)}{2}\] \[b_n-1=n(n-1)\] \[b_n=n(n-1)+1\]

Now we can substitute $a_n$ back into our equation:

\[n(a_n-1)=n(n-1)+1\] \[a_n-1=n-1+\frac{1}{n}\] \[a_n=n+\frac{1}{n}\] \[a_n^2=n^2+\frac{1}{n^2}+2\]

Thus the sum becomes

\[\sum^{100}_{n=1} a_n^2= \sum^{100}_{n=1} n^2+ \sum^{100}_{n=1} \frac{1}{n^2}+ \sum^{100}_{n=1} 2\]

We know that $\sum^{100}_{n=1} n^2=\frac{100\cdot101\cdot201}{6}=338350$, and we also know that $\sum^{100}_{n=1} 2=200$, so the requested sum is equivalent to $\sum^{100}_{n=1} \frac{1}{n^2}+338550$. All that remains is to calculate $\sum^{100}_{n=1} \frac{1}{n^2}$, and we know that this value lies between $1$ and $2$ (see the note below for a proof). Thus,

\[\sum^{100}_{n=1} a_n^2= \sum^{100}_{n=1} \frac{1}{n^2}+338550<2+338550=338552\] \[\sum^{100}_{n=1} a_n^2= \sum^{100}_{n=1} \frac{1}{n^2}+338550>1+338550=338551\]

so

\[\sum^{100}_{n=1} a_n^2\in(338551,338552)\]

and thus the answer is $\boxed{\textbf{(B) }338551}$.

~eevee9406

See also

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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
All AMC 12 Problems and Solutions

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