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Difference between revisions of "1992 AHSME Problems/Problem 21"

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<cmath>= a_1 + (a_1 + a_2) + (a_1 + a_2 + a_3) + \ldots + (a_1 + a_2 + \ldots + a_n)</cmath>
 
<cmath>= a_1 + (a_1 + a_2) + (a_1 + a_2 + a_3) + \ldots + (a_1 + a_2 + \ldots + a_n)</cmath>
 
<cmath>= n \cdot a_1 + (n - 1) \cdot a_2 + \ldots + a_n.</cmath>
 
<cmath>= n \cdot a_1 + (n - 1) \cdot a_2 + \ldots + a_n.</cmath>
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If we take this to be the Cesáro total for the second sequence, we can see that all the terms but the first term make up the first sequence. Since we know that the Cesáro total of the original sequence is <math>1000 * 99 = 99,000,</math> than the Cesáro total of the second sequence is <math>n \cdot a_1 + 99,000 = 100 * 1 + 99,000 = 99,100.</math> Thus the Cesáro sum of the second sequence is <math>\frac{99,100}{100} = \boxed{991, A}\, .</math>
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If we take this to be the Cesáro total for the second sequence, we can see that all the terms but the first term make up the first sequence. Since we know that the Cesáro total of the original sequence is <math>1000 * 99 = 99,000,</math> than the Cesáro total of the second sequence is <math>n \cdot a_1 + 99,000 = 100 \cdot 1 + 99,000 = 99,100.</math> Thus the Cesáro sum of the second sequence is <math>\frac{99,100}{100} = \boxed{991, A}\, .</math>
  
 
== See also ==
 
== See also ==

Revision as of 16:18, 18 August 2020

Problem

For a finite sequence $A=(a_1,a_2,...,a_n)$ of numbers, the Cesáro sum of A is defined to be $\frac{S_1+\cdots+S_n}{n}$ , where $S_k=a_1+\cdots+a_k$ and $1\leq k\leq n$. If the Cesáro sum of the 99-term sequence $(a_1,...,a_{99})$ is 1000, what is the Cesáro sum of the 100-term sequence $(1,a_1,...,a_{99})$?

$\text{(A) } 991\quad \text{(B) } 999\quad \text{(C) } 1000\quad \text{(D) } 1001\quad \text{(E) } 1009$

Solution

Let us define the Cesáro total of a particular sequence to be n * Cesáro sum. We can see that the Cesáro total is \[S_1 + S_2 + S_3 + ... + S_n\] \[= a_1 + (a_1 + a_2) + (a_1 + a_2 + a_3) + \ldots + (a_1 + a_2 + \ldots + a_n)\] \[= n \cdot a_1 + (n - 1) \cdot a_2 + \ldots + a_n.\] If we take this to be the Cesáro total for the second sequence, we can see that all the terms but the first term make up the first sequence. Since we know that the Cesáro total of the original sequence is $1000 * 99 = 99,000,$ than the Cesáro total of the second sequence is $n \cdot a_1 + 99,000 = 100 \cdot 1 + 99,000 = 99,100.$ Thus the Cesáro sum of the second sequence is $\frac{99,100}{100} = \boxed{991, A}\, .$

See also

1992 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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All AHSME Problems and Solutions

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