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

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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.

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What is sum formulea Of this type sequence

I got nth term formulea and sum formulea

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...

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I got suitable formulea for nth term and sum this type sequence and more sequence

1+2+4+7+11+16+22.......

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.

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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.

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This is formula regarding to error method

I have exact formulae

But first I will regist it

1+2+4+7+11+16+........

Cool my all Maths expert

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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 .....

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

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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

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1+0=1 1+1=2 1+3=4 ...

general rule is :

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

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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

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