f(x)=ax2+bx+cf(1)=a+b+c=3Now f(x+y)=f(x)+f(y)+xyput y=1f(x+1)=f(x)+f(1)+xf(x+1)+x+3Nowf(2)=7f(3)=12NowSn=3+7+12+.....tn...........(1)Sn=3+7+.....tn−1+tn.........(2)On subtracting (2) from (1)tn=3+4+5+.... upto n termstn=2n2+5nSn∑tn=∑2n2+5nSn=21[6n(n+1)(2n+1)+25n(n+1)]S10=330