Вправи до параграфів: §.17 (5)

5. Знайдіть y_n−1 + y_n+1, якщо:

y_n–1 + y_n+1 – ?

1) y_n = n + 3.

y_n = yn-1+ yn+12

y_n–1 + y_n+1 = 2y_n = 2(n + 3) = 2n + 6.

2) y_n = 5n2

y_n–1 + y_n+1 = 5n2 • 2 = 5n.

3) Довести:

y_n = 6 – 2n.

y_{n – 1} = 6 – 2(n – 1) = 6 – 2n + 2 = 8 – 2n.

y_{n + 1} = 6 – 2(n + 1) = 6 – 2n – 2 = 4 – 2n.

y_n = yn+1+ yn-12 = 4-2n+8-2n2 = 6 – 2n.

Є арифм. прогресією.

4) y_n = 5n-23

y_n–1 = 5n-1- 23 = 5n-5-23 = 5n-73

y_n+1 = 5n+1- 23 = 5n+5-23 = 5n+33

y_n = yn+1+ yn-12 = 5n+33+ 5n-732 = 5n+3+5n-76 = 10n-46 = 5n-23

Є арифм. пр.