jmc

Question

For a finite sequence A=(a_1,a_2,,a_n) of numbers, the Cesaro sum of A is defined to be S_1++S_nn,where S_k=a_1++a_k and 1 k n. If the Cesaro sum of the 99-term sequence (a_1,,a_99) is 1000, what is the Cesaro sum of the 100-term sequence (1,a_1,,a_99)?

Accepted Answer

Applying the definition to the sequence

(a_{1}, a_{2}, \dots, a_{99}),

we get

\frac{a _{1} + ( a _{1} + a _{2} ) + \dots + ( a _{1} + a _{2} + \dots + a _{99} )}{99} = 1000.

Thus,

99 a_{1} + 98 a_{2} + \dots + 2 a_{98} + a_{99} = 99000.

Then the Cesaro sum of

(1, a_{1}, a_{2}, \dots, a_{99})

is

\frac{1 + ( 1 + a _{1} ) + ( 1 + a _{1} + a _{2} ) + \dots + ( 1 + a _{1} + a _{2} + \dots + a _{99} )}{100} = \frac{100 + 99 a _{1} + 98 a _{2} + \dots + 2 a _{98} + a _{99}}{100} = \frac{100 + 99000}{100} = \frac{99100}{100} = 991 .