Competition problem — Algebra — Maths Olympiad Prep

Library / /257 of 520

Algebra Difficulty 3.1 AMC 10/12 Find the answer

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
$(1,a_1,...,a_{99})$ ?

Pick one

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 $\frac{99,100}{100} = \boxed{991, A}\, .$

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.