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Question

Find the sum of series up to n terms:

131+13+231+3+13+23+331+3+5+....

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Solution

The nth term of the given series is

13+23+33+...+n31+3+5+...+(2n1)=[n(n+1)2]21+3+5+...+(2n1)

Here ,1,3,5,...(2n1) is an A.P. with first term a , last term (2n1) and number of terms as n

1+3+5+...+(2n1)=n2[2×1+(n1)2]=n2

an=n2(n+1)24n2=(n+1)24=14n2+12n+14

Sn=nk=1ak=nk=1(14K2+12K+14)

By using ni=1k=1+2+3++n=n(n+1)2 and

ni=1k2=12+22+32++n2=n(n+1)(2n+1)6 we have,

=14n(n+1)(2n+1)6+12n(n+1)2+14n

=n[(n+1)(2n+1)+6(n+1)6]24

=n[2n2+3n+1+6n+6+6]24

=n(2n2+9n+13)24


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