Question

zadnai od 7.44 do 7.48 prosze pomocy ​

Answer 1

Verified answer

Odpowiedź:

7.44

a) r=3

[tex]-2,1,4.7,10,13,16[/tex]

b) r=3

[tex]-6,-3,0,3,6,9,12[/tex]

c) r=-2

[tex]3,1,-1,-3,-5,-7,-9[/tex]

d) r=3/2

[tex]\frac{1}{2} ,2,\frac{2}{7} ,2,5,\frac{13}{2} ,8,\frac{19}{2}[/tex]

7,45

a)

[tex]a_{1}=7\\a_{2}=3\\r=a_{2}-a_{1}=3-7=-4\\a_{n}=a_{1}+(n-1)*r\\a_{n}=7+(n-1)*(-4)\\a_{n}=7-4n+4\\a_{n}=-4n+11[/tex]

b)

[tex]a_{1}=-1\\a_{2}=4\\r=a_{2}-a_{1}=4-(-1)=4+1=5\\a_{n}=-1+(n-1)*5\\a_{n}=-1+5n-5\\a_{n}=5n-6[/tex]

c)

[tex]a_{1}=-3\sqrt{2} \\a_{2}=\sqrt{2} \\r=\sqrt{2} +3\sqrt{2} =4\sqrt{2} \\a_{n}=-3\sqrt{2} +(n-1)*4\sqrt{2} \\a_{n}=-3\sqrt{2} +4\sqrt{2} n-4\sqrt{2} \\a_{n}=4\sqrt{2} n-7\sqrt{2}[/tex]

d)

[tex]a_{1}=-\sqrt{5} \\a_{2}=0\\r=0-(-\sqrt{5}) =\sqrt{5} \\a_{n}=-\sqrt{5} +(n-1)*\sqrt{5} \\a_{n}=-\sqrt{5} +\sqrt{5}n -\sqrt{5} \\a_{n}=\sqrt{5} n-2\sqrt{5}[/tex]

7,46

a)

[tex]a_{n}=4n-3\\a_{1}=4*1-3=1\\a_{n+1}=4(n+1)-3=4n+4-3=4n+1\\r_{a_{n+1}-a_{n} }=4n+1-(4n-3)=4n+1-4n+3=4[/tex]

b)

[tex]a_{n} =5n+2\\a_{1} =5*1+2=7\\a_{n+1}=5*(n+1)+2=5n+5+2=5n+7\\r=5n+7-(5n+2)=5n+7-5n-2=5[/tex]

c)

[tex]a_{n}=8-5n\\a_{1}=8-5*1=3\\a_{n+1}=8-5*(n+1)=8-5n-5=-5n+3\\r=-5n+3-(8-5n)=-5n+3-8+5n=-5[/tex]

d)

[tex]a_{n}=\sqrt{3} -\sqrt{2}n\\a_{1}=\sqrt{3} -\sqrt{2} *1=\sqrt{3} -\sqrt{2}\\a_{n+1}=\sqrt{3} -\sqrt{2} (n+1)=\sqrt{3} -\sqrt{2} n-\sqrt{2} \\r=\sqrt{3} -\sqrt{2} n-\sqrt{2}-(\sqrt{3} -\sqrt{2}n)=\sqrt{3} -\sqrt{2} n-\sqrt{2}-\sqrt{3} +\sqrt{2}n=-\sqrt{2}[/tex]

7,47

                                   [tex]a_{n} -a_{n-1}=a_{n+1}-a_{n}[/tex]

a) TAK

[tex]a_{n}=4n+3\\a_{n-1}=4(n-1)+3=4n-4+3=4n-1\\a_{n+1}=4(n+1)+3=4n+4+3=4n+7\\4n+3-(4n-1)=4n+7-(4n+3)\\4n+3-4n+1=4n+7-4n-3\\4=4[/tex]

b) TAK

[tex]a_{n}=n+\sqrt{2} \\a_{n-1}=n-1+\sqrt{2} \\a_{n+1}=n+1+\sqrt{2} \\n+\sqrt{2} -(n-1+\sqrt{2})=n+1+\sqrt{2}-(n+\sqrt{2})\\n+\sqrt{2} -n+1-\sqrt{2}=n+1+\sqrt{2}-n-\sqrt{2}\\1=1[/tex]

c) NIE

[tex]a_{n} =\frac{1}{3} *(8n-3)\\a_{n-1}=\frac{1}{3}* (8*(n-1)-3)=\frac{1}{3} *(8n-8-3)=\frac{1}{3} *(8n-11)=\frac{8}{3}n-\frac{11}{3} \\\\a_{n+1}=\frac{1}{3}* (8*(n+1)-3)=\frac{1}{3} *(8n+8-3)=\frac{1}{3} *(8n+5)=\frac{8}{3}n +\frac{5}{3} \\\\\frac{1}{3} *(8n-3)-(\frac{8}{3}n-\frac{11}{3})=\frac{8}{3}n +\frac{5}{3}-(\frac{1}{3} *(8n-3))\\\\\frac{8}{3}n-1 -\frac{8}{3}n+\frac{11}{3}=\frac{8}{3}n +\frac{5}{3}-\frac{8}{3}n +1\\\\-1+\frac{11}{3} =1+\frac{5}{3} \\\\[/tex]

d) TAK

[tex]a_{n} =n-1\\a_{n-1}=n-1-1=n-2\\a_{n+1}=n+1-1=n\\n-1-(n-2)=n-(n-1)\\n-1-n+2=n-n+1\\1=1[/tex]

e) NIE

[tex]a_{n}=n^{2} \\a_{n-1}=(n-1)^2=n^2-2n+1\\a_{n+1}=(n+1)^2=n^2+2n+1\\n^2-(n^2-2n+1)=n^2+2n+1-(n^2)\\n^2-n^2+2n-1=n^2+2n+1-n^2\\2n-1=2n+1[/tex]

f) NIE

[tex]a_{n}=(n-1)^2\\a_{n-1}=(n-1-1)^2=(n-2)^2=n^2-4n-4\\a_{n+1}=(n+1-1)^2=n^2\\(n-1)^2-(n^2-4n-4)=n^2-(n-1)^2\\n^2-2n+1-n^2+4n+4=n^2-(n^2-2n+1)\\n^2-2n+1-n^2+4n+4=n^2-n^2+2n-1\\2n+5=2n-1[/tex]

7,48

a)                                     [tex]a_{n} =a_{1} +(n-1)*r[/tex]

[tex]a_{3} =a_{1} +(3-1)*1\\10 =a_{1} +2\\-a_{1} =2-10\\-a_{1}=-8/(-1)\\a_{1}=8[/tex]

[tex]a_{7} =a_{1} +(7-1)*1\\a_{7}=8+6\\a_{7}=14[/tex]

b)

[tex]a_{9} =a_{1}+(9-1)*5\\4=a_{1}+40\\-a_{1}=40-4\\-a_{1}=36/(-1)\\a_{1}=-36[/tex]

[tex]a_{7}=a_{1}+(7-1)*5\\a_{7}=-36+30\\a_{7}=-6[/tex]

c)

[tex]a_{2}=a_{1}+(2-1)*2\\5=a_{1}+2\\-a_{1}=2-5\\-a_{1}=-3/(-1)\\a_{1}=3[/tex]

[tex]a_{7}=a_{1}+(7-1)*2\\a_{7}=3+12\\a_{7}=15[/tex]

d)

[tex]a_{6}=a_{1}+(6-1)*(-3)\\-2=a_{1}-18\\-a_{1}=-18+2\\-a_{1}=-16/(-1)\\a_{1}=16[/tex]

[tex]a_{7}=a_{1}+(7-1)*(-3)\\a_{7}=16-18\\a_{7}=-2[/tex]

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