# Answer Part (b)?

Mar 22, 2018

See below.

#### Explanation:

If

${a}_{n} = 2 {n}^{2} - 3 n + 5$ then

${a}_{n + 1} = 2 {\left(n + 1\right)}^{2} - 3 \left(n + 1\right) + 5 = 2 {n}^{2} + 4 n + 2 - 3 n - 3 + 5 = {a}_{n} + 4 n - 1 = 4 \left(n + 1\right) - 5$

but we know that

${a}_{n} = {a}_{n - 1} + 4 n - 5$

and we have shown that for $n \ge 2$

${a}_{n - 1} = {a}_{n} + 4 \left(n + 1\right) - 5$

The final arrangements for the finite induction demonstration are left to the reader.