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Matematika / Ketma-ketliklar va binom

Rekurrent ketma-ketliklar

Test: 10 ta savol, vaqt bilan

4 ta oson, 4 ta oʻrta, 2 ta qiyin. Tez va toʻgʻri javob — koʻproq ball.

Testni boshlash

Yoki quyida mashq qiling: variantni tanlang — javob va izoh chiqadi. Ball berilmaydi.

Oson · 5 ta

1.

a1=4a_1 = 4, an+1=−2an+5a_{n+1} = -2a_n + 5. a5a_{5} ni toping.

2.

a1=5a_1 = 5, an+1=an+na_{n+1} = a_n + n. a5a_{5} ni toping.

3.

a1=−2a_1 = -2, an+1=−2an−4a_{n+1} = -2a_n - 4. a5a_{5} ni toping.

4.

a1=−3a_1 = -3, an+1=an+na_{n+1} = a_n + n. a6a_{6} ni toping.

5.

a1=−1a_1 = -1, an+1=−2an+3a_{n+1} = -2a_n + 3. a4a_{4} ni toping.

Oʻrta · 5 ta

1.

a1=−1a_1 = -1, an+1=an2−1a_{n+1} = a_n^2 - 1. a3a_{3} ni toping.

2.

a1=4a_1 = 4, a2=3a_2 = 3, an+2=an+1+ana_{n+2} = a_{n+1} + a_n. a6a_{6} ni toping.

3.

a1=2a_1 = 2, an+1=an2−3a_{n+1} = a_n^2 - 3. a4a_{4} ni toping.

4.

a1=1a_1 = 1, a2=1a_2 = 1, an+2=an+1+ana_{n+2} = a_{n+1} + a_n. a7a_{7} ni toping.

5.

a1=−1a_1 = -1, an+1=an2−3a_{n+1} = a_n^2 - 3. a4a_{4} ni toping.

Qiyin · 5 ta

1.

a1=−5a_1 = -5, an+1=12 an−4a_{n+1} = \frac{1}{2}\,a_n - 4. Ketma-ketlikning limitini toping.

2.

a1=5a_1 = 5, an+1=2an−4a_{n+1} = 2a_n - 4. a5a_{5} ni toping.

3.

a1=0a_1 = 0, an+1=12 an−6a_{n+1} = \frac{1}{2}\,a_n - 6. Ketma-ketlikning limitini toping.

4.

a1=−1a_1 = -1, an+1=2an−2a_{n+1} = 2a_n - 2. a4a_{4} ni toping.

5.

a1=−1a_1 = -1, an+1=14 an+5a_{n+1} = \frac{1}{4}\,a_n + 5. Ketma-ketlikning limitini toping.

Yordam