1. ## Induction

Verify by induction that the solution of recurrence:

$\displaystyle F(1) = 1$
$\displaystyle F(n) = F(n-1) +3n+2$

is $\displaystyle F(n) = \frac{(3n^2)}{2} + \frac{7n}{2-4}$

2. Originally Posted by Apprentice123
Verify by induction that the solution of recurrence:

$\displaystyle F(1) = 1$
$\displaystyle F(n) = F(n-1) +3n+2$

is $\displaystyle F(n) = \frac{(3n^2)}{2} + \frac{7n}{2-4}$
2- 4= -2 do you mean $\displaystyle F(n)= \frac{3n^2}{2}- \frac{7n}{2}= \frac{3n^2- 7n}{2}$?

3. Originally Posted by HallsofIvy
2- 4= -2 do you mean $\displaystyle F(n)= \frac{3n^2}{2}- \frac{7n}{2}= \frac{3n^2- 7n}{2}$?
Sory the correct is $\displaystyle F(n) = \frac{3n^2}{2} + \frac{7n}{2} -4$

4. Originally Posted by Apprentice123
Sory the correct is $\displaystyle F(n) = \frac{3n^2}{2} + \frac{7n}{2} -4$
While it can be difficult to solve equations, it can be much easier to verify a solution. With $\displaystyle F(n)= \frac{3n^2}{2}+ \frac{7n}{2}- 4$, $\displaystyle F(n-1)= \frac{3(n-1)^2}{2}+ \frac{7(n-1)}{2}- 4= \frac{3n^2- 6n+ 3+ 7n- 7- 8}{2}= \frac{3n^2+ n- 12}{2}$.

Then $\displaystyle F(n-1)+ 3n+ 2= \frac{3n^2+ n- 12}{2}+ 3n+ 2$$\displaystyle = \frac{3n^2+ n- 12}+ \frac{6n+ 4}{2}= \frac{3n^2+ 7n- 8}{2}= \frac{3n^3}{2}+ \frac{7n}{2}- 4= F(n)$.

5. Thank you