Question

Find an expression for the general term of the series. Assume the starting value of the index, k is 1.

Answer 1

\[\frac{ (x-a) }{ 1 } + \frac{ (x-a)^2 }{ 3*2! } + \frac{ (x-a)^3 }{ 9*3! } + \frac{ (x-a)^4 }{ 27*4! } + . . .\]

Answer 2

looks like the denominator has a term that looks like \(3^k\) but if you want to start at \(k=1\) then you have to make that part \(3^{k-1}\)

Answer 3

then next part would be \(k!\)

Answer 4

general term would look like
\[\frac{(x-a)^k}{3^{k-1}k!}\]

Answer 5

thank you so much