Answer :
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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