Answer:
Hello,
Step-by-step explanation:
41)
[tex]2,4,6,8,10,...\\\\u_0=2 , u_{n+1}=u_n+2\\\\u_n=2*n+2\\\\[/tex]
42)
[tex]0,6,12,18,24,...\\\\u_0=0 , u_{n+1}=u_n+6\\\\u_n=6*n+0\\\\[/tex]
43)
[tex]-5,-4,-3,-2,-1,...\\\\u_0=-5 , u_{n+1}=u_n+1\\\\u_n=1*n-5=n-5\\\\[/tex]
44)
[tex]-4,-8,-12,-16,-20,...\\\\u_0=-4 , u_{n+1}=u_n-4\\\\u_n=-4*n-4\\\\[/tex]
45)
[tex]-5,-3.5,-2,-0.5,1,...\\\\u_0=-5 , u_{n+1}=u_n+1.5\\\\u_n=1.5*n-5\\\\[/tex]
46)
[tex]-32,-20,-8,4,16,...\\\\u_0=-32 , u_{n+1}=u_n+8\\\\u_n=8*n-32\\\\[/tex]
48)
[tex]0,\dfrac{1}{8},\dfrac{2}{8},\dfrac{3}{8},...\\\\u_0=0 , u_{n+1}=u_n+\dfrac{1}{8}\\\\u_n=\dfrac{1}{8}*n+0\\\\[/tex]
49)
[tex]27,15,3,-9,-21,...\\\\u_0=27 , u_{n+1}=u_n-12\\\\u_n=-12*n+27\\\\[/tex]
47) be carefull this is not a arithmetic sequence but a quadratic.
I show you the general method.
[tex]\begin{array}{c|c|c|c}x&\Delta_1&\Delta_2&\Delta_3\\--&--&--&--\\1&.&.&.\\&&&\\\dfrac{11}{3} &\dfrac{8}{3} &.&.\\&&&\\4 &\dfrac{1}{3}&\dfrac{-7}{3} &.\\&&&\\2 &-2 &\dfrac{-7}{3}&0\\&&&\\--&--&--&--\\\end {array} \\\\Since\ \Delta_3\ is\ null,\ u_n=a*n^2+b*n+c\ has\ for\ degree\ 2.\\[/tex]
[tex]\begin {array} {c|c|c|c|c}n&u_n&equation&\\--&--&-----&\\0&1&=a*0+b*0+c &c=1\\1&\dfrac{11}{3}&=a*1+b*1+c&a+b=\dfrac{8}{3} \\2&4&=a*4+b*2+c&4a+2b=3\\\end {array}\\\\\\\left\{\begin {array} {ccc}4a+2b&=&3\\a+b&=&\dfrac{8}{3}\\\end {array} \right.\\\\\\\left\{\begin {array} {ccc}a&=&\dfrac{-7}{6}\\\\b&=&\dfrac{23}{6}\\\\c&=&1\\\\\end {array} \right.\\\\\\\boxed{u_n=\dfrac{-7}{6}*n^2+\dfrac{23}{6}*n+1} \\[/tex]
[tex]u_n=\dfrac{-7}{6}*n^2+\dfrac{23}{6}*n+1}\\\\u_{n+1}=\dfrac{-7}{6}*(n+1)^2+\dfrac{23}{6}*(n+1)+1}\\\\u_{n+2}=\dfrac{-7}{6}*(n+2)^2+\dfrac{23}{6}*(n+2)+1}\\\\\Delta_1(n)=u_{n+1}-u_{n}=\dfrac{-7}{6}*(2n+1)+\dfrac{23}{6}\\\Delta_1(n+1)=u_{n+2}-u_{n+1}=\dfrac{-7}{6}*(2n+3)+\dfrac{23}{6}\\\\\Delta_2(n)=\Delta_1(n+1)-\Delta_1(n)=\dfrac{-7}{6}*2 \\[/tex]
[tex]\left\{\begin{array}{ccc}u_{n+2}-2u_{n+1}+u_n&=&\dfrac{-7}{3} \\\\u_0&=&1\\\\u_1&=&\dfrac{11}{3} \end {array} \right.[/tex]
