S1=1, S2=[2,3], S3=[4,5,6], S4=[7,8,9,0]. How do i find 1st term of general Sn?

Question

Ans= [n(n-1)/2] + 1

Answer 1

first term in S1 = 1 = 0 + 1 = ∑0 + 1
first term in S2 = 2 = 1 + 1 = ∑1 + 1
first term in S3 = 4 = 3 + 1 = ∑2 + 1
first term in S4 = 7 = 6 + 1 = ∑3 + 1
first term in S5 = 11 = 10 + 1 = ∑4 + 1
first term in S6 = 16 = 15 + 1 = ∑5 + 1
first term in S7 = 22 = 21 + 1 = ∑6 + 1
first term in S8 = 29 = 28 + 1 = ∑7 + 1
..................................................................
whence
first term in Sn = ∑(n--1) + 1 = (n--1)*n/2 + 1

Answer 2

S1: 1 number
S2: 2 numbers
S3: 3 numbers
S4: 4 numbers
....
S(n-1): (n-1) numbers
S(n):

1+2 +3 + 4 +... + (n -1) = (n-1) *n/2
So, the first number in S(n) will be (n-1) *n/2+ 1

Answer 3

You gave the answer, more or less, although it's a little unclear since you made typos.

There are at least two major ways to prove that the sum of the first n natural numbers is n(n+1)/2. One way is by induction. Add n+1 to n(n+1)/2 and you get (n+1)(n+2)/2. That's the heart of the inductive proof.

The other is to line up the numbers 1 + 2 + 3 + ... + 100 (say) over the numbers 100 + ... + 1, and notice that you now have 100 columns, each of which adds to 101. So 100*101 is twice the sum.

Answer 4

n...........A
1..........1
2...........2
3............4
4...........7
clearly this is a sequence with parabolic shape
use an^2+bx+c=A
a+b+c=1
4a+2b+c=2
9a+3b+c=4
solve for a,b,c
a=1/2
b=-1/2
c=1
therefore general is
A=n^2/2-n/2+1