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)






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