Question

Express the sequence {a_n}_n = 1^infinity as an equivalent sequence of the form {b_n}_n = 3^infinity {n^2 + 7n - 5}_n = 1^infinity An equivalent sequence is { }_n = 3^infinity (Simplify your answer. DO not factor.)

Answer 1

We're given the sequence,

`\{b_n\}_(n=3)^\infty=\{25,39,55,73,\ldots\}`

Since
`b_n` is quadratic,
`a_n` should also be quadratic. Replace
`n` with
`n-2` (if
`n=3`, then
`n-2=1`) in the definition of
`b_n`, and let
`a,b,c` denote the new coefficients:

`n^2+7n-5=a(n-2)^2+b(n-2)+c`

Expand the right side:

`n^2+7n-5=an^2+(-4a+b)n+(4a-2b+c)`

Coefficients of terms with the same degree should be the same:

`\begin{cases}a=1\\-4a+b=7\\4a-2b+c=-5\end{cases}\implies a=1,b=11,c=13`

So the new sequence is the same, with

`\{a_n\}_(n=1)^\infty=\{25,39,55,73,\ldots\}`