can you help with my maths?

Question

can anyone help me find the nth term for this
3 8 17 30 47

Answer 1

First let me start off by saying, I see the pattern but do not know how to express it for the nth term.

3 + 5 = 8
8 + 9 = 17
17 + 13 = 30
30 + 17 =47
47
I would have to say the pattern is the initial distance is 5 than plus 4 on the next iteration.

Answer 2

Find the nth term for these numbers

3 8 17 30 47

3 + 5 = 8

8 + 9 = 17

17 + 13 = 47

47 + 21 = 68

The nth term is 68

Answer 3

Tn = 2n^2 + 11n - 5

Term 2: 3+5=8

Term 3: 8+(5+4)=17
=> 3+5+(5+4)=17 => 3+5(2)+4=17

Term 4: 17+(5+4+4)=30
=> 3+5+(5+4)+(5+4+4)=30 => 3+5(3)+4(sum of 1 and 2)=30

Term 5: 30+(5+4+4+4)=47
=> 3+5+(5+4)+(5+4+4)+(5+4+4+4)=47
=> 3+3(5)+4(sum of 1 to 3)=47

Term n:
3 + 5(n-1) + 4(sum of 1 to n) + [5+4(n-2)]
= 3 + 5n - 5 + 4[(n/2)(2+(n-1)] + 5 + 4n - 8
= 3 + 5n - 5 + 4n + 2n^2 - 2n + 5 + 4n - 8
= 2n^2 + 11n - 5

So, Tn = 2n^2 + 11n - 5.

NB: The sum of 1 to n is a arithmetic progression, so the formula I used was Sn = (n/2)[2a+(n-1)d] where a is the first term, d is the difference and n is the number of terms.

Answer 4

Right, u need to find the difference between the numbers, if its not constant look at second difference
3......8.........17.......30........47
5 9 13 17 1st diff
4 4 4 2nd diff

Because we have a second diff we have an x^2 (x squared) and because the common diff is 4, we have 2x^2, so thats part of our sequence.

Then we look at what we add on to each term
for first term 1^2=1 1x2+1=3
2^2=4 4x2+0 =8
3^2=9 2x9-1=17
4^2=16 2x16-2=30
so we look at the sequence 1...........0...........-1............-1
-1 -1 -1

so we -n+2 for this part of the sequence so....

your nth term is 2n^2-n+2 - Hope that makes sense to you, its hard to explain online!

Answer 5

68
LOOK:
3+ _=8 =>5
8+_=17 =>9
17+_=30 =>13
30+_=47=>17

5+4=9+4=13+4=17
17+4=21=>21+47=68

Answer 6

3,8,17,30,47

first difference

5,9,13,17,21?,25?,...........

second difference

4,4,4,4,........?

3+5 = 8
3+5+(5+4)=17
3+5+(5+4)+(5+4+4) =30 ha ha the penny has dropped
3+5+(5+4)+(5+4+4)+(5+4+4+4) =47
3+5+(5+4)+(5+4+4)+(5+4+4+4)
+(5+4+4+4+4)= 68 -i've got the drift

therefore, we have

Nn =3+5n+2n(n-1)

= 2n^2 + 3(n+1)

n=0,1,2,...............

=2(n^2+1) -n (changing n limits)

n=1,2,3,........................

where Nn = the nth term

since it was a second difference,the formula has to be in the form of a second order polynomial

i hope this helps

Answer 7

yes, the Nth term is:
2n²-n+2
because it goes down to the second difference.

Answer 8

If you mean what is the next number I am guessing 68, if not then I haven't a clue.