Difference between revisions of "1992 AHSME Problems/Problem 21"
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Let us define the Cesáro total of a particular sequence to be n * Cesáro sum. We can see that the Cesáro total is | Let us define the Cesáro total of a particular sequence to be n * Cesáro sum. We can see that the Cesáro total is | ||
| − | + | <cmath>S_1 + S_2 + S_3 + ... + S_n | |
| − | 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.</cmath> | |
| − | |||
| − | |||
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> | 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> | ||
Revision as of 16:17, 18 August 2020
Problem
For a finite sequence 
of numbers, the Cesáro sum of A is defined to be
, where 
and 
. If the Cesáro sum of
the 99-term sequence 
is 1000, what is the Cesáro sum of the 100-term sequence
?
Solution
Let us define the Cesáro total of a particular sequence to be n * Cesáro sum. We can see that the Cesáro total is
![\[S_1 + S_2 + S_3 + ... + S_n = a_1 + (a_1 + a_2) + (a_1 + a_2 + a_3) + \ldots + (a_1 + a_2 + \ldots + a_n) = n \cdot a_1 + (n - 1) \cdot a_2 + \ldots + a_n.\]](http://latex.artofproblemsolving.com/b/b/a/bbac8574ca9f9cbfa58e6d1c760688d51eeaf899.png)
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 
than the Cesáro total of the second sequence is 
Thus the Cesáro sum of the second sequence is 
See also
| 1992 AHSME (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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
