What is the Nth Term for this sequence 1,1,9,9,25?

Question

The sequence is 1, 1, 9, 9, 25

I've been told that there isn't a definite Nth term but if you can find one that would be very helpful. Working will be appreciated.

Also could you try these sequences too:

1, 5, 13, 25, 41 (Total No. Of Squares)

0, 4, 8, 12, 16 (Border Squares)

4, 24, 60, 112, 180 (Right Angles)

Again working would be appreciated. The Names in brackets are just so I can remember which sequences they are.

Answer 1

1, 1, 9, 9, 25
For an odd n, the nth term is n^2.
For an even n, the nth term is (n-1)^2.

Answer 2

Its a not difficulty-free one, yet i'm able that ought to help you: (Sum(i) from i =a million to i = n) = n(n+a million)/2 Examples: (Sum(i) from i = a million to i = 5) = a million+2+3+4+5 = (5)(6)/2 = 15 (Sum(i) from i =a million to i=10) = a million+2+3+4+5+6+7+8+9+10 = (10)(eleven)/2 = fifty 5 The equation (n)(n+a million)/2 supplies us the effect of the summation without us having to characteristic up each and every of the numbers. you want the equation for the sum of the even numbers. we can get this with the help of multiplying the i with the help of two and then factoring out both. (Sum(2i) from i=a million to i=n) = 2(Sum(i) from i=a million to i=n) =2n(n+a million)/2 = n(n+a million) party: sum(2i) from i=a million to i=5)= 2+4+6+8+10 = (5)(6) = 30 you've performed the paintings to finish that you sequence must be written: 2 + (sum(2i) from i=a million to n) = 2 + n(n+a million) = n^2 + n + 2 If we call both the 0th time period of the sequence, 4 the first, 8 the 2d and so on the equation n^2 + n + 2 works tremendous. we may rather call 2 the1st time period contained in the sequence, 4 the 2d, 8 the 3th and so on. No biggie, lets make cause them to act like they are one a lot less with the help of plugging in n-a million for n. We get: (n-a million)^2 + (n-a million) + 2 = n^2 - 2n + a million +n -a million + 2 = n^2 - n + 2 the reply on your question is n^2 - n + 2. are you able to describe why that's continually real that: (Sum(i) from i =a million to i = n) = n(n+a million)/2 We under no circumstances proved it. when you're intense about math you may favor to provide it a attempt. imagine about averages. What ought to the overall of a set a million+2+3+4+5+..n equivalent?

Answer 3

the first one doesn't even seem to be a function

for 1,5,13,25,41 seems to be

n^2+(n-1)^2

for 0, 4, 8, 12, 16 seems to be

(n-1)*4

for 4, 24, 60, 112, 180 seems to be

4n(2n-1)

Answer 4

a) f(n) = n^2 if n is odd
= (n-1)^2 if n is even

or to get rid of the even odd part:
f(n) = [n-0.5 - 0.5cos(pi*n)]^2

b) f(n) = f(n-1) + (n-1)*4 where f(1) = 1
Leons is also valid for this

c) f(n) = 4*(n-1)

d) f(n) = 4n(2n-1)

Answer 5

not enogh time
need to work and make money to live

Answer 6

its go like this 1^2,1^2,3^2,3^2,5^2,5^2,7^2,7^2.........the ninth position is 7^2
So the answer is 49

Answer 7

(n+2)squared

I think