solve for the 10th term or nth term of the sequence ?

let us look at

1,3,7,15,31,63, 127 ...

the nth term = 2^n- 1

now let us look at

1, (3,7), ( 15.31, 63) ...

the nth set has got n elements

before 1st element of nth set there are n(n-1)/2 elements

so 1st element of nth set is (n-1)(n)/2 + 1 = (n^2-n + 2)/2

in (1) , (3 + 7) , (15 + 31 + 63) , (127 + 255 + 511 + 1023) , _ _ _

so nth term sum has 1st term = 2^ ((n^2-n + 2)/2) - 1

there are n terms in each sum

so sum = 2^ ((n^2-n + 2)/2) ( 1 + 2 + 2^2. .. + 2^(n-1) - n

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

the first term in the sequence can be rewritten as 2^1 - 1 = 1

the second term can be rewritten as (2^2 -1) + (2^3 - 1) = 3 + 7 = 10

for the third term: (2^4 - 1) + (2^5 - 1) + (2^6 - 1) = 15 + 31 + 63 = 109

and so on...

in general, the nth term of this sequence has the form

[2^(k+1) - 1] + [2^(k+2) - 1] + ... + [2^(k+n) - 1]

where k is the sum of the first (n-1) natural numbers, for which the

explicit formula [(n-1)^2 + (n-1)]/2 can be used.

Therefore, for the 10th term of the series you have first to compute the sum

of the first 9 natural numbers, which is given by (9^2 + 9)/2 = 45,

after which you know that the term you are looking for is

(2^46 -1) + (2^47 - 1) + ... + (2^55 - 1)

this is a very big number, and I think it is better to write a computer program that make

the required calculations for you.

P.S. the above formula can be simplified even further, using the fact that the

equality 2^1 + 2^2 + ... 2^n = 2^(n+1) -1 for any n greater or equal to 0.

Then you can rewrite the 10th term of the series above as

(2^1 + ... + 2^45) + (2^46 + ... + 2^55) -

(2^1 + ... 2^45) -

10 =

(2^56) - 1 + (2^46 - 1) - 10 = 2^56 - 2^46 - 12

What does the sequence go up in? It is going up in 3's so which you place 3n, then you truly would desire to artwork out what the huge style earlier the 7 could have been, so eliminate 3 and it is 4, so the nth term is: 3n + 4

I may not know the answer, but I can see the pattern...

A1 = 2-1

A2 = (4-1) + (8-1)

A3 = (16-1) + (32-1) + (64-1)

A4 = (128-1) + (256-1) + (512-1) + (1024-1)

see it?

A1 = (2^n - 1)

A2 = (2^n - 1) + (2^(n+1)-1)

A3 = (2^(n+1)-1) + (2^(n+2)-1) + (2^(n+3)-1)

A4 = (2^(n+3)-1) + (2^(n+4)-1)...........

I can't find out how to transfer that to A10 or An, but Surely You've learned how?