In the nth row of the triangle                        1               2            3        4            5             67             8            9            10………………………. ……………………….

Last term of nth row =1+2+3+....+n=12n(n+1) As terms in the nth row forms an A.P. with common difference 1,first  term=last  term−(n−1)(1)=12n(n+1)−n+1=12(n2−n+2)Sum   =12n[12(n2−n+2)+12(n2+n)]=12n(n2+1).