The sum of the terms in the nth bracket of the series (1)+(2+3+4)+(5+6+7+8+9)+…,is

For n=1, we haveSum of the terms in first bracket = 1 and, (n−1)3+n3=(1−1)3+13=1For n = 2, we have3um of the terms in the second bracket = 2 + 3 + 4 = 9 and, (n−1)3+n3=(2−1)3+23=1+8=9For n = 3, we haveSum of the terms in the third bracket = (5 + 6 + 7 + 8 + 9) =35and, (3−1)3+33=8+27=35Hence, option (a) is true.