It is nor an
AP or
GPLet Sn=3+7+13+21+31+....+an−1+an ----- ( 1 )
Sn=0+3+7+13+21+...+an−2+an−1+an ----- ( 2 )
Subtracting ( 2 ) from ( 1 )
Sn−Sn=3−0[(7−3)+(13−7)+(21−13)+....+(an−1−an−2)+(an−an−1)]−an
0=3+[4+6+8+....+an−1]−an
an=3+[4+6+8+....+an−1] ----- ( 3 )
Now, 4+6+8+....+an−1 is an AP
Whose first term a=4
Common difference d=6−4=2
We know that,
Sum of n terms of AP=n2(2a+(n−1)d)
Putting, n=n−1,a=4,d=2
[4+6+8+...(n−1)terms]=(n−12)[2a+(n−1−1)d]
=(n−12)[2(4)+(n−2)2]
=(n−12)[8+2n−4]
=(n−12)(2n+4)
=(n−12)2(n+2)
=(n−1)(n+2)
Thus, [4+6+8...upto(n−1)terms]=(n−1)(n+2)
From ( 3 )
an=3+[4+6+8+...+an−1]
Putting values
=3+(n−1)(n+2)
=3+n(n+2)−1(n+2)
=3+n2+2n−n−2
=n2+n+3−2
=n2+n+1
Now,
Sn=n∑n=1an
=n∑n=1n2+n+1
=n∑n=1n2+n∑n=1n+n∑n=11
=n(n+1)(2n+1)6+n(n+1)2+n
=n(n+1)(2n+1)+3n(n+1)+6n6
=n((n+1)(2n+1)+3(n+1)+66)
=n(2n2+n+2n+1+3n+3+66)
=n(2n2+6n+106)
=n6×2(n2+3n+5)
=n3(n2+3n+5)
∴ The required sum is n3(n2+3n+5)