# Sum the following

Let s=4 + 6 + 9 + 13 + 18 + …………. + n

shifting each term by one,

s=4 + 6 + 9 + 13 + 18 + ……………….. + nth …..(1)

s= 4 + 6 + 9 + 13 + 18 + …………. + (n - 1)th + nth …(2)

by (1) - (2) we get

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

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

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

Nth = 4 + (summation upto (n - 1)th term)

we know

Substituting the above-given value in (3)

thus

s = 4 + 6 + 9 + 13 + 18 + …………. + nth =

s = 4 + 6 + 9 + 13 + 18 + …………. + 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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