help plz!!!?

Question

write the first five terms of the seqeunce whose nth term is:
n2-5
2n2 (the 2 is sqaured by the way)
most informative answer gets my 10 points

Answer 1

All you do is substitute the numbers 1 to 5 into the sequence as the first five terms means n=1..5

so n^2- 5 is:

1^2 - 5 = -4
2^2-5 = -1
3^2-5 = 4
4^2-5=11
5^2-5=20

Similary with 2*n^2 gives 2,8,18,32,50

Answer 2

If you mean that the nth term is n squared -5 (also written as
n^2-5), then the first 5 terms are:
1^2 -5 = 1 - 5 = -4
2^2-5= 4-5 = -1
3^2-5 = 9-5 = 4
4^2-5 = 16-5 = 11
5^2-5= 25-5 = 20

Answer 3

don't worry these are a piece of cake.....

n can represent any number...in this case it's 1,2,3,4,5,

the first equation means ( as long as it's not squared and it does mean two? if it's squared then look at the other version)

n
1x2-5 = -3
2x2-5 = -1
3x2-5 = 1
4x2-5 = 3
5x2-5 = 5

if it's n2 - 5 (squared)

1x1-5 = -4
2x2-5 = -1
3x3-5 = 4
4x4-5 = 11
5x5-5 = 20


for the second equation

2x (n x n) (do the barcket first)

2x(1x1) = 2
2x(2x2) = 8
2x(3x3) = 18
2x(4x4) = 32
2x(5x5) = 50

that takes me back, i haven't done those since gcse's good stuff - that was in 1998!! - i'm 24 yrs old now!!!! god that makes me feel old :(

Answer 4

let N=nth term
for n^2-5,
n=1,N= -4
n=2,N= -1
n=3,N=4
n=4,N=11
n=5,N=20
n=n,N=n^2-5

sequence: -4,-1,4,11,20,...(n^2-5)

1st difference 3,5,7,9
2nd difference 2,2,2,

we want a formula in the form of
an^2+bn+c
N=an^2+bn+c
-4=a+b+c, (n=1)
-1=4a+2b+c, (n=2)
4=9a+3b+c, (n=3)
sweep
3=3a+b
5=5a+b
2a=2 >>>>a=1,
substitute back
b=0,c= -5
N=n^2+0n-5
=n^2-5 as required

since this is a second difference
sequence,the formula for N is a second
order polynomial,namely
n^2-5

now for 2n^2
n=1,N=2
n=2,N=8
n=3,N=18
n=4,N=32
n=5,N=50
n=n,N=2n^2

sequence:2,8,18,32,50,..(2n^2)

1st difference 6,10,14,18
2nd difference 4, 4, 4,

we want a formula in the form of
an^2+bn+c
N=an^2+bn+c
2=a+b+c, (n=1)
8=4a+2b+c, (n=2)
18=9a+3b+c, (n=3)
sweep
6=3a+b
10=5a+b
2a=4>>>>a=2
substitute back
b=0,c=0
N=2n^2+0n+0
= 2n^2 as required

since this is a second difference
sequence,the formula for N is a second
order polynomial,namely
2n^2

i hope that this increases your
knowledge of numerical analysis

Answer 5

You need to re-write your question so it is more clear. The way you have this written now, it's impossible to tell whether this is a linear progression or a geometric sequence. I surmise that it is probably geometric, but I doubt you will receive many responses until you do this.

Answer 6

-4 , -1, 4 , 9 , 20 ,31 45

(1) - (2) - (3) (4) (5) (6 (7)


1*1 = 1 - 5 = -4

and so on

Answer 7

substitute n=1,2,3,4,5 and get the corresponding values
(1)^2-5
(2)^2-5
(3)^2-5
4)62-5
(5)^2-5
-4,-1,4,11,20
(1)^2
(2)^2
(3)^2
(4)^2
(5)^2
2,8,18,32,50

Answer 8

Sadly I no longer have to do homework so I'm definitley not doing yours!

Answer 9

i hate questions with n as a variable.. i'm better with the x and y questions

Answer 10

are we doing your homework for you, or are you just curious?