Sum the following series to n terms :
1 + 3 + 6 + 10 + 15 + …………….
Let s = 1 + 3 + 6 + 10 + 15 + …………. + n
By shifting each term by one
S = 1 + 3 + 6 + 10 + 15 + …………. + nth …………… (1)
S = 1 + 3 + 6 + 10 + …………. + (n - 1)th + nth ….(2)
by (1) - (2) we get
0 = 1 + (2 + 3 + 4 + 5 + …….nth - (n - 1)th - n)
Nth = 1 + (2 + 3 + 4 + 5 + …….nth - (n - 1)th - nth)
Nth = 1 + (2 + 3 + 4 + …….r + 1) ……………(3)
Nth = 1 + (summation upto (n - 1)th term)
we know
Substituting the above-given value in (3)
thus
S =3 + 5 + 9 + 15 + 23 + …………. + nth =
We know by property that:
∑axn + bxn - 1 + cxn - 2…….d0=a∑xn + b∑xn - 1 + c∑xn - 2…….. + d0∑1
…… (4)
We know
Thus substituting the above values in(4)