Hello, fortuna!

Here's #3 . . .

The characteristic equation is: .t^n .= .10·t^{n-1} - 25·t^{n-2}3. Solve the recurrence relation: .an .= .10·an-1 - 25·an-2, .a0 = 2, a1 = 15.

Hint: If the characteristic equation factors to (t - r)²,

. . . . then the general solution is: .an .= .b·r^n + c·n·r^n

. . Divide by t^{n-2}: . t² .= .10t - 25 . . → . . t² - 10t + 25 .= .0

. . Hence: .(t - 5)² .= .0

Then the general solution is: .an .= .b·5^n + c·n·5^n

Since a0 = 2, we have: .b·5^0 + c·0·5^0 .= .2 . . → . . b = 2

Since a1 = 15 and b = 2, we have: .2·5 + c·1·5 .= .15 . . → . . c = 1

Therefore, the general solution is: .an .= .2·5^n + n·5^n