Q. 63.7( 3 Votes )
Sum the following series to n terms :
1 + 4 + 13 + 40 + 121 + ……….
Answer :
Let s=1 + 4 + 13 + 40 + 121 + …………. + n
By shifting each term by one
S = 1 + 4 + 13 + 40 + 121 + …………. + nth ….(1)
S = 1 + 4 + 13 + 40 + …………. + (n - 1)th + nth …(2)
by (1) - (2) we get
0 = 1 + (3 + 9 + 27 + 81 + …….nth - (n - 1)th - n)
Nth = 1 + (3 + 32 + 33 + 34 + …….nth - (n - 1)th - nth)
Nth = 1 + (3 + 32 + 33 + …….3n - 1) ………(3)
we know
Substituting the above-given value in (3)
thus
s = 1 + 4 + 13 + 40 + 121 + …………. + 
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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