For a finite sequence $A=(a_1,a_2,\dots,a_n)$ of numbers, the Cesaro 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 Cesaro sum of the 99-term sequence $(a_1,\dots,a_{99})$ is 1000,  what is the Cesaro sum of the 100-term sequence $(1,a_1,\dots,a_{99})$?
Now, when I insert a 1 at the beginning of the sequence, the new partial sums are $S'_k=1+S_{k-1}$ for $1\leq k\leq 100$, where I define $S_0=0$. Therefore, the Cesaro sum of the new 100-term sequence is \[\frac{S'_1+\cdots+S'_{100}}{100}.\] This means that I can write the Cesaro sum of the new sequence as \[\frac{1+(1+S_1)+\cdots+(1+S_{99})}{100}=\frac{100+(S_1+\cdots+S_{99})}{100}.\] Now, I can use the fact that the Cesaro sum of the original sequence is 1000, which implies that \[S_1+\cdots+S_{99}=1000\cdot 99.\] Substituting this into the expression for the Cesaro sum of the new sequence, I get \[\frac{100+(S_1+\cdots+S_{99})}{100}=\frac{100+1000\cdot 99}{100}=\frac{99100}{100}=991.\]

