Zadanie wykonam w załączniku
5) Sn=a1(1-g^n) / 1-g a1(1-2^4) / 1-2=60 a1*(-15) =-60 -15a1= -60 /÷(-15) a1=60/15 a1=4 4) a1+3r=6 /*(-1) a1+7r=14 -a1-3r= -6 a1+7r=14 4r=8 /÷4 r=2 a1 +7r=14 a1+7*2=14 a1=0 a10=a1+9r=0+9*2=18 S10=a1+an /2 * n= 0+18 /2 * 10=90 3) a2=a1+3 a3=a1+6 Tw. Pitagorasa; a1² +a2² =a3² a1² +(a1+3)² =(a1+6)² a1² +a1² +6a1 +9 =a1² +12a1 +36 a1² -6a1 -27 √Δ=36-4*1*(-27)=36+108=√144=12 a1=6-12 /2= -3 (Odpada bo jest minusowe (-)) a1=6+12 /2=9 a1=9 a2=a1+3=9+3=12 a3=a1+6=9+6=15 Obw.=9+12+15=36 2) an=1/2(n-3) an+1 =1/2(n+1-3)= 1/2(n-2) an+1 - an= 1/2(n-2) - 1/2(n-3)= 1/2n -2/2 -1/2n +3/2=1/2 r=1/2 WIELKOŚĆ STAŁA TO JEST CIĄG ARYTM. 1) an+1 = n+2 /2n+3= n+1+2 / 2(n+1)+3 = n+3 / 2n+2+3 = n+3 / 2n+5 an+1 - an = n+3 / 2n+5 - n+2 / 2n+3 = (n+3)(2n+3) - (n+2)(2n+5) / (2n+5)(2n+3)= 2n²+3n+6n+9 -(2n² +5n+4n+10) / (2n+5)(2n+3) = 2n² +9n+9 - 2n² -9n-10 / (2n+5)(2n+3) = - 1 / (2n+5)(2n+3) < 0 ciąg malejący b) an+1= n² -n= (n+1)² - (n+1)= n²+2n+1-n-1=n² +n an+1 - an= n²+n - (n²-n) = n²+n-n² +n= 2n > 0 ciąg rosnący
