13.5+15.7+17.9+…to n terms =
n3(2n+3)
n2n+3
1(n+2)(n+4)
none of these
For n=1, have
n3(2n+3)=135=First term of the series
For n=2, we have
n3(2n+3)=23.7=135+15.7= Sum of first two terms of the series
Hence, option (a) is correct
ALITER We have
135+15.7+17.9+to n terms
=12235+25.7+279+…+2(2n+1)(2n+3)=1213−15+15−17+17−19+…+12n+1−12n+3=1213+12n+3=n3(2n+3)