Let s=1 + 3 + 7 + 13 + 21 + …………. + n

By shifting each term by one

s=1 + 3 + 7 + 13 + 21 + …………. + nth (1)

s= 1 + 3 + 7 + 13 + …………. + (n - 1)th + nth (2)

by (1) - (2) we get

0=1 + 2 + 4 + 6 + 8 + …….nth - (n - 1)th - nth

nth=1 + (2 + 4 + 6 + 8 + …….2r ) (3)

nth=1 + (summation of first (n - 1)th term)

we know

Substituting the above given value in (3)

nth=1 + n² - n

general term =1 + r² - r

thus

s=1 + 3 + 7 + 13 + 21 + …………. + 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)