Find the sum of n terms of the series 3+8+22+72+266+1036+...........

tn=a*n(square)+b*n+c t1=a+b+c=3---(1) t2=4a+2b+c=8----(2) t3=9a+3b+c=22----(3) on solving (1),(2),(3) we get a=9/2;b=-17/2;c=7; now general term of the sereis is: tn=(9/2)*n(square)-(17/2)*n+7 sn=(9/2)*n(n+1)(2n+1)/6-(17/2)*n(n+1)/2+7n is the soln of the question

anwer is incorrect pls give me correct soln.... ans..is 1/3(4^n-1)+n(n+1)