Question

Write an explicit and a recursive formula for each sequence.

41. 2, 4, 6, 8, 10....
42. 0, 6, 12, 18, 24,...
43.-5.-4,-3,-2.-1...
44. -4,-8,-12,-16, -20,...
45. - 5,-3.5,-2,-0.5,1,...
46.-32,-20,-8, 4, 16, ...
47. 1,11/3,12/3,2,...
48. 0,1/8,1/4,3/8,...
49. 27, 15, 3,-9, -21....​

Answer 1

41. 12, 14

42. 30, 36

43. 0, 1

44. -24, -28

45. 2.5 , 4

46.

47.

48.

49. 63 , 189

Explanation:

I don't know answers to 46 , 47 , 48

please make my answer as brainelist

Answer 2

Hello,

Explanation:

41)

`2,4,6,8,10,...\\\\u_0=2 , u_(n+1)=u_n+2\\\\u_n=2*n+2\\\\`

42)

`0,6,12,18,24,...\\\\u_0=0 , u_(n+1)=u_n+6\\\\u_n=6*n+0\\\\`

43)

`-5,-4,-3,-2,-1,...\\\\u_0=-5 , u_(n+1)=u_n+1\\\\u_n=1*n-5=n-5\\\\`

44)

`-4,-8,-12,-16,-20,...\\\\u_0=-4 , u_(n+1)=u_n-4\\\\u_n=-4*n-4\\\\`

45)

`-5,-3.5,-2,-0.5,1,...\\\\u_0=-5 , u_(n+1)=u_n+1.5\\\\u_n=1.5*n-5\\\\`

46)

`-32,-20,-8,4,16,...\\\\u_0=-32 , u_(n+1)=u_n+8\\\\u_n=8*n-32\\\\`

48)

`0,(1)/(8),(2)/(8),(3)/(8),...\\\\u_0=0 , u_(n+1)=u_n+(1)/(8)\\\\u_n=(1)/(8)*n+0\\\\`

49)

`27,15,3,-9,-21,...\\\\u_0=27 , u_(n+1)=u_n-12\\\\u_n=-12*n+27\\\\`

47) be carefull this is not a arithmetic sequence but a quadratic.

I show you the general method.

`\begin{array}cx&\Delta_1&\Delta_2&\Delta_3\\--&--&--&--\\1&.&.&.\\&&&\\(11)/(3) &(8)/(3) &.&.\\&&&\\4 &(1)/(3)&(-7)/(3) &.\\&&&\\2 &-2 &(-7)/(3)&0\\&&&\\--&--&--&--\\\end {array} \\\\Since\ \Delta_3\ is\ null,\ u_n=a*n^2+b*n+c\ has\ for\ degree\ 2.\\`

`\begin {array} cn&u_n&equation&\\--&--&-----&\\0&1&=a*0+b*0+c &c=1\\1&(11)/(3)&=a*1+b*1+c&a+b=(8)/(3) \\2&4&=a*4+b*2+c&4a+2b=3\\\end {array}\\\\\\\left\{\begin {array} {ccc}4a+2b&=&3\\a+b&=&(8)/(3)\\\end {array} \right.\\\\\\\left\{\begin {array} {ccc}a&=&(-7)/(6)\\\\b&=&(23)/(6)\\\\c&=&1\\\\\end {array} \right.\\\\\\\boxed{u_n=(-7)/(6)*n^2+(23)/(6)*n+1} \\`

`u_n=(-7)/(6)*n^2+(23)/(6)*n+1}\\\\u_(n+1)=(-7)/(6)*(n+1)^2+(23)/(6)*(n+1)+1}\\\\u_(n+2)=(-7)/(6)*(n+2)^2+(23)/(6)*(n+2)+1}\\\\\Delta_1(n)=u_(n+1)-u_(n)=(-7)/(6)*(2n+1)+(23)/(6)\\\Delta_1(n+1)=u_(n+2)-u_(n+1)=(-7)/(6)*(2n+3)+(23)/(6)\\\\\Delta_2(n)=\Delta_1(n+1)-\Delta_1(n)=(-7)/(6)*2 \\`

`\left\{\begin{array}{ccc}u_(n+2)-2u_(n+1)+u_n&=&(-7)/(3) \\\\u_0&=&1\\\\u_1&=&(11)/(3) \end {array} \right.`