If 1413+2435+345.7+…+n4(2n−1)(2n+1)=148f(n)+n16(2n+1),  then f (n) is equal to

We have,tn=n4(2n−1)(2n+1)=n24+116+1164n2−1=4n2+116+13212n−1−12n+1∴Sn=∑n=1n tn=∑n=1n 4n2+116+132∑n=1n 12n−1−12n+1=14n(n+1)(2n+1)6+116n+1321−12n+1=n484n2+6n+5+1321−12n+1=n4n2+6n+548+n16(2n+1)∴f(n)=n4n2+6n+5