Let S = 2 + 4 + 7 + 11 + 16 + …………. + n

By shifting each term by one

S = 2 + 4 + 7 + 11 + 16 + ……………….. + nth ….(1)

S = 2 + 4 + 7 + 11 + 16 + …………. + (n - 1)th + nth …(2)

by (1) - (2) we get

0 = 2 + (2 + 3 + 4 + 5 + …….nth - (n - 1)th - nth)

Nth = 2 + (2 + 3 + 4 + 5 + …….nth - (n - 1)th)

nth = 2 + (2 + 3 + 4 + …….r + 1) …………(3)

nth = 2 + (summation upto (n - 1)th term)

we know

Substituting the above-given value in (3)

thus

s=2 + 4 + 7 + 11 + 16 + …………. + nth =

s=2 + 4 + 7 + 11 + …………. + nth =

We know by property that:

∑axⁿ + bx^{n - 1} + cx^{n - 2}…….d₀=a∑xⁿ + b∑x^{n - 1} + c∑x^{n - 2}…….. + d₀∑1

….(4)

We know

Thus substituting the above values in(4)