if 1 + 2 + 3 + . . . + n = n ( n + 1 ) , then 3 ( 1 + 3 + 5 + . . . . + 69 ) = ?
explanation : to solve this use the formula of ap , sn = ( n / 2 ) ( a + l ) . . . . . . . . . . . . . . . . ( 1 ) to find n , use = > tn = a + ( n - 1 ) d = > 69 = 1 + ( n - 1 ) 2 = > n = 35 use value of n in ( 1 ) then , sn = ( 35 / 2 ) ( 1 + 69 ) = 1225 ans : - 3 ( sn ) = 3675 answer : a
multiply(subtract(divide(multiply(69, add(69, const_1)), const_2), multiply(divide(subtract(69, const_1), const_2), add(divide(subtract(69, const_1), const_2), const_1))), const_3)