# What is the general term of 1,2,4,7,11,16,22?

1. What is the general term of 1,2,4,7,11,16,22?

Is there any technique how to find the general term? plsss help me.

1. Addition of 1 2 3 4 5 6 7......
1+1=2
2+2=4
3+4=7
4+7=11
5+11=16
and so on

1. The Fibonacci Sequence is the series of numbers: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ... The next number is found by adding up the two numbers before it.

2. It is not fibonacci series. Fibonacci series is like 0,1,1,2,3,5,8,13...... (adding adjacent number).
The peculiarity of the question raised ( general term 1,2,4,7,11,16,22). Adding natural number to each of the terms in the series, we get the result.

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2. I'm not sure if this is a mathematical question, but it looks like you have posted a series of numbers that flow in the following way:
1+1 = 2
2+2 = 4
4+3 = 7
7+4 = 11
11+5 = 16
16+6 = 22

1 was added to the 1st number to get 2, that sum, 2, was added to 2, because it was the second number, to get 4.  The sum of 4 was added to 3, because it was the third number, to get 7.  Yada, yada, yada...

1. I got suitable formulea for nth term and sum this type sequence and more sequence
1+2+4+7+11+16+22.......

3. yah, but what is the exact formula to come up with this kind of problem.. thank you for your answer. ^^

4. To find the nth term of the sequence:

0.5*n^2 - 0.5*n + 1

If we look at the difference between successive terms, we see that the differences are not the same - so a linear function will not work.

1  2  3  4  5  6

But if we look at the difference between successive differences (second-order differences, if you will), we see that these are the same - so a quadratic function will work.

1  1  1  1  1  1

To find the function, I looked for some pattern.  I listed the triangular numbers (1 3 6 10 15 21 27); the nth triangular number is the sum of the positive integers less than or equal to n.  It appears as though the nth term of the sequence is adjusted down (by differing amounts) from the triangular number.  Specifically, to find the nth term, we subtract (n-1) from the nth triangular number.

This yields:

(Sum of numbers 1 through n) - (n-1)

= 0.5(n)(n+1) - n + 1

= 0.5*n^2 + 0.5*n - n + 1

= 0.5*n^2 - 0.5*n + 1

I know that I've merely given you my thought process rather than a hard, fast method to be applied every time.  All I can say is that finding the nth term is like solving any other problem type: you get better with experience.  You build an intuition.

I hope this helps.

1. This is formula regarding to error method

5. I have exact formulae
But first I will regist it
1+2+4+7+11+16+........
Cool my all Maths expert

6. I would call this a sequential equation e.g. 1,2,4,7,11,16,22, 29.  You see:
1+1=2
2+2=4
4+3=7
7+4=11
11+5=16
16+6=22
22+7=29.....and 29+8=37, 37+9=46, the main component is adding sequential numbers to the equation.

7. The simple formula of this number series is like that:-

1st Digit + 1= 2nd Digit,
2nd Digit + 2= 3rd Digit.
3rd digit + 3= 4th Digit.

So, Nth Digit = (N-1)th Digit + (N-1): It is the formula.

So, if n=5, then the 5th Digit = (5-1)th Digit+(5-1)
= 4th Digit + 4
= 7+4= 11.

1. What is sum formulea Of this type sequence
I got nth term formulea and sum formulea

8. 1+1=2
2+(1+1)=4
4+(1+2)=7
7+(1+3)=11
11+(1+4)=16
16+(1+5)=22
22+(1+6)=29

9. Xn = X(n-1) + (n-1) is the formula you are looking for.

Where Xn is the nth (or general) term of the sequence.
So, X(n-1) is the term directly before Xn.
And (n-1) is just the integral value of n minus 1.

Example: n=5
Xn = 11
X(n-1) = 7
n-1 = 5-1 = 4

So using the formula, 11 = 7 + 4

10. 1+0=1   1+1=2   1+3=4    ...

general rule is :

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

11. 1 2 4 7 11 16 22

1st difference:  0 1 2 3 4 5 6 7.........

1+0=1
1+1=2
2+2=4
4+3=7
7+4=11
11+5=16
16+6= 22
22+7= 29

And so .....

12. next comes 29
Each time you add up one more than the last time.
So  to the first no ie:  1  you add 1

13. Yes there is a technique it is called Quadratic Sequence

1,2,4,7, 11, 16,22

Difference 1 2 3 4 5 6
0+1=1
1+1=2
2+2=4
3+4=7
4+7=11
5+11=16
6+16=22
7+22=29
8+29=37
And so on

14. 1,2,4,7,11,16,22 next one is 29. check it once http://bit.ly/2pHf8sk

15. next is 29 get solution here https://bit.ly/3MaYBTn

16. Communications link failure by Mulesoft 4.x only in localhost

I'm using mulesoft 4.x. I have the following problem: I can't get it to connect to mysql server in localhost, but I can get it to connect on remote hosts without problem. However, that server "localhost" works correctly so much that it is reachable by the mulesoft runtime if I move it to another machine.

the error is: Caused by: org.mule.runtime.extension.api.exception.ModuleException: java.sql.SQLException: Cannot get connection for URL jdbc:mysql://localhost:3306/test?logger=org.mule.extension.db.api.logger.MuleMySqlLogger : Communications link failure

Example: VM with ip 10.0.2.200 and mulesoft runtime installed with MYSQL. mulesoft does not connect, but if I move mulesoft to a different server with ip 10.0.2.202, it connects to Mysql on 10.0.2.200