Help for calculating this series....please help have an exam tomorrow.....

how can i calculate the sum or find the nth term of this kinda series...
1+4+8+13+19+26+33....
here,the common differences are serially this 3,4,5,6,7,....this forms a sequence of common difference of 1.....how can calculate the nth term or the summation up to nth term of this series???please help.....

It looks like a quadratic sequence. If it is, then we should be able to say:
nth term = an² + bn + c

1st term = 1, so
a + b + c = 1 …… eqn 1

2nd term = 4, so
4a + 2b + c = 4 …… eqn 2

3rd term = 8, so
9a + 3b + c = 8 …… eqn 3

Three equations, three unknowns. Here we go! Eqn 2 – eqn 1:
4a + 2b + c = 4 –
a + b + c = 1
–––––––––––––
3a + b = 3 …… eqn 4

Eqn 3 – 3(eqn 4):
9a + 3b + c = 8 –
9a + 3b = 9
–––––––––––––
c = -1

Into eqn 1:
a + b = 2

Compare with eqn 4:
2a = 1
a = 1/2

b = 3/2

SOLUTION: Here's the formula for the sequence: nth term = n²/2 + 3n/2 - 1
It might be easier as: nth term = n(n + 3)/2 - 1

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I'm still trying to find where Jacinta provided the formula for the summation to term n, which was also part of your question. It must be there somewhere for you to have awarded her 5 stars... : )

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Note that S=1+4+8+13+19+26+33...+t_k
S= 1+4+8+13+19+26.....+t_k-1+t_k
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(Subtracting) 0=1+[3+4+5+6+...upto (k-1)th term]-t_k
=>t_k=1+(k-4)/2*[2*3+k-4]
=>t_k=(K^2-2K)/2-3

T(1) = 1
T(n) = T(n-1) + n + 1 = (n+1)C2 + (n-1) = (n^2 + n) / 2 + n - 1 = n^2 / 2 + 3n/2 -1

S(1) = 1
S(n) = n^2 + (n+1)C3 = n^2 + (n^3 - n) / 6 = (n^3 + 6n^2 - n) / 6