Question

Help please? Finding sequence

Answer 1

`1)`

`a_1 = 1`
`a_2 = 1^2+1 = 2`
`a_3 = 2^2+1 = 5`
`a_4 = 3^2+1 = 10`
`a_5 = 4^2 + 1 = 17`

`2)`

`a_1 = 1^2 + 2 xx 1 = 3`
`a_2 = 2^2 + 2 xx 2 = 8`
`a_3 = 3^2 + 2 xx 3 = 15`
`a_4 = 4^2 + 2 xx 4 = 24`
`a_5 = 5^2 + 2 xx 5 = 35`

Answer 2

`"see explanation"`

Explanation:

`(1)`

`"this is a "color(blue)"recurrence relationship"`

`"where each term is obtained by substituting the "`
`"previous term into the equation"`

`"the first term is given, that is "a_1=color(red)(1)`

`rArra_2=(color(red)(1))^2+1=1+1=color(blue)(2)`

`rArra_3=(color(blue)(2))^2+1=4+1=color(magenta)(5)`

`rArra_4=(color(magenta)(5))^2+1=25+1=color(purple)(26)`

`a_5=(color(purple)(26))^2+1=676+1=677`

`"the first 5 terms are "1,2,5,26,677`

`(2)`

`"obtain the terms in this sequence by substituting"`
`"1,2,3,4 and 5 into the n th term formula"`

`rArra_1=1^2-(2xx1)=1-2=-1`

`a_2=2^2-(2xx2)=4-4=0`

`a_3=3^2-(2xx3)=9-6=3`

`a_4=4^2-(2xx4)=16-8=8`

`a_5=5^2-(2xx5)=25-10=15`

`"the first 5 terms are "-1,0,3,8,15`

Answer 3

`color(red)((1))1,2,5,26,677.`
`color(red)((2))-1,0,3,8,15.`

Explanation:

`color(blue)((1) a_1=1), and`

`color(red)(a_n=(a_(n-1))^2+1`

`a_2=(a_(2-1))^2+1=(a_1)^2+1=(1)^2+1=1+1=2`

`a_3=(a_(3-1))^2+1=(a_2)^2+1=(2)^2+1=4+1=5`

`a_4=(a_(4-1))^2+1=(a_3)^2+1=(5)^2+1=25+1=26`

`a_5=(a_(5-1))^2+1=(a_4)^2+1=(26)^2+1=676+1=677`

Hence. first five terms :`1,2,5,26,677.`

`(2)`

`a_n=n^2-2n`

`a_1=(1)^2-2(1)=1-2=-1`

`a_2=(2)^2-2(2)=4-4=0`

`a_3=(3)^2-2(3)=9-6=3`

`a_4=(4)^2-2(4)=16-8=8`

`a_5=(5)^2-2(5)=25-10=15`

Hence. first five terms : `-1,0,3,8,15.`