What is the Nth term of this sequence: 1, 2, 4, 7, 11?

Question

Please Help Me If You Know!

x

Answer 1

come on now, 1+1=2,2+2=4,4+3=7,7+4=11... 11+5=16...

Answer 2

16...

Answer 3

try this... starting with the n=0:

1 + (n+1)n/2

Answer 4

T(n) = n(n-1)/2 + 1

= (n^2 - n + 2) / 2

Answer 5

yea its 16.

1 (+1)=2

2 (+2)=4

4 (+3)= 7

7+ (4)=11

11 (+5)= 16

16 (+6)=22

Answer 6

the 1st order diffrences 1,2,3,4

so it is a second order polynomial and the term t we can put as

xt = A + B(t-1) + C(t-1)(t-2)

( I can chose any oplynomial od degree 0 1 2 etc to be independent and I choose above form for easy computaion)

put t = 1to get A = 1

xt = 1 + B(t-1) + C(t-1)(t-2)



put t = 2 to get B+1 = 2 or B= 1

xt = 1 + 1(t-1) + C(t-1)(t-2)

put t =3 i get

4 = 1 + 2 + 2c OR C = 1/2

So

xt = 1 + 1(t-1) +1/2(t-1)(t-2)

= 1/2(t-1)(t-2) + t

= 1/2(t^2- 3t +2 +2t)

= 1/2(t^2-t + 2)

math kp

Answer 7

I understand the question, but I haven't been able to figure out the formula.  It is, however, driving me nuts!!!



you're getting some "interesting" answers though.

Answer 8

You always do a +1 to the value you added with before. So the answer is 11 + 5 = 16

Hope I've helped =)

Answer 9

OK Ms F,

you have a function that uses RECURSION.  The nth element has to refer to the previous element, just like the famous fibonacci sequence.

see http://en.wikipedia.org/wiki/Fibonacci_n...

For your sequence, the function F is 1 for the first element and then F(n-1) plus (n-1).  Thus F(3) = F(2) + 2 = 4,

F(4) = F(3) + 3 = 7 etc as show above.

The nth element is F(n-1) + n-1.

Answer 10

1+0=1

1+1=2

2+2=4

4+3=7

7+4=11

11+5=16

16+6=22

22+7=29

29+8=37

37+9=46