z.9 a) a1 = 1, a2 = 1/2, a3 = 3,a4 = 1/4, a5 = 5,a6 = 1/6 b) a1 = 1^2 = 1 a2 = (-1)^2 * 2^2 = 1*4 = 4 a3 = 3^2 = 9 a4 = (-1)^4 *4^2 = 1*16 = 16 a5 = 5^2 = 25 a6 = (-1)^6 * 6^2 = 1*36 = 36 c) a1 = - 1 a2 = 2^3 = 2*2*2 = 8 a3 = - 3 a4 = 4^3 = 64 a5 = - 5 a6 = 6^3 = 36*6 = 216 ===================== z.14 a) an = n^2 - 6n + 5 n^2 - 6n + 5 = 0 delta = 36 -4*1*5 = 36 - 20 = 16 n = [6 -4]/2 = 2/2 = 1 lub n = [ 6 + 4]/2 = 10/2 = 5 Odp. a1 = 0 i a5 = 0 ===================== b) an = n*(n^2 - 1) = n*( n -1)*(n + 1) an = 0 <=> n = 1 Odp.a1 = 0 ============== c) an = [ n^2 - 3n + 2]/( n + 1) an = 0 <=> n^2 -3n + 2 = 0 delta = 9 -4*1*2 = 9 - 8 = 1 n = [3 - 1]/2 = 2/2 = 1 lub n = [3 + 1]/2 = 4/2 = 2 zatem a1 = 0 oraz a2 = 0 ========================== d) an = [ n^3 - 9n^2 + 14 n ]/[n^2 + 4 ] an = 0 <=> n^3 - 9n^2 +14 n = 0 <=> n*(n^2 - 9n + 14) = 0 <=> <=> n^2 - 9n + 14 = 0 delta = 81 - 4*1*14 = 81 - 56 = 25 n = [9 - 5]/2 = 4/2 = 2 lub n = [ 9 + 5 ]/2 = 14/2 = 7 Odp. a2 = 0 oraz a7 = 0 ========================= e) an = (1/n)(n^2 - 3)(n^2 - 1) an = 0 <=> n = 1 Odp. a1 = 0 ================== f) an = p(n + 1) - 4/ p(n + 1) an = p(n +1) - [ 4*p(n +1)]/(n + 1) an = [(n + 1) p(n +1)]/(n + 1) - [4 p(n +1)]/(n +1) an = [ p(n +1)*( n + 1 - 4)]/( n + 1) an = [ p(n +1)*(n - 3)]/ (n + 1) = 0 <=> n - 3 = 0 <=> n = 3 Odp.a3 = 0 =============== z.15 a) an = n^2 - 2n - 14 an = 1 <=> n^2 - 2n - 14 = 1 <=> n^2 - 2n - 15 = 0 delta = 4 - 4*1*(-15) = 4 + 60 = 64 n = [ 2 - 8]/2 = -6/2 = - 3 < 0 - odpada, bo n musi być liczbą naturalną lub n = [ 2 + 8 ]/2 = 10/2 = 5 Odp.a5 = 1 ==================== b) an = [ n^2 - 6n + 15 ]/(n + 3) an = 1 <=> n^2 - 6n + 15 = n + 3 <=> n^2 - 7n + 12 = 0 delta = (-7)^2 - 4*1*12 = 49 - 48 = 1 n = [ 7 - 1]/2 = 6/2 = 3 lub n = [ 7 + 1]/2 = 8/2 = 4 Odp. a3 = 1 oraz a4 = 1 ========================== c) an = ( -1)^n an = 1 dla n parzystych, czyli a2 = a4 = a6 = a8 = .... = 1 ============================== z.15 a) an = 10 - n^2; m = 0 an > 0 <=> 10- n^2 > 0 <=> 10 > n^2 <=> n^2 < 10 n = 1 lub n = 2 lub n = 3 Odp. a1 > 0, a2 > 0, a3 > 0 ============================ b) an = 2^n - 6; m = 10 an > 10 <=> 2^n - 6 > 10 <=> 2^n > 16 <=> n > 4 Odp. a5 > 10, a6 > 10, a7 > 10, a8 > 10, .... ============================================= c) an = ( 3 - n)/ (2 n); m = 0 an > 0 <=> (3 - n)/(2n) > 0 < => 3 - n > 0 <=> n < 3 n = 1 lub n = 2 Odp. a1 > 0, a2 > 0 ==================== d) an = ( 3n - 3)/( n + 1) ; m = 2 an > 2 <=> (3n - 3)/(n + 1) > 2 <=> 3n - 3 > 2*(n + 1) <=> 3n - 3 > 2n + 2 <=> <=> n > 5 Odp. a6 >2, a7 > 2, a8 > 2, ... ==============================
Ciągi!!! Prosze o wytłumaczenie także ,bo jakoś ciągi nie są dla mnie zbyt jasne:) z Góry dziekuje:)
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