JEE Mains Sequence and Series 2

It's probably easiest to calculate the complement first.

The sum of the integers between $1$ and $100$ inclusive that are divisible by $3$ is

$\displaystyle\sum_{k=0}^{33} (3k) = 3*\dfrac{33*34}{2} = 1683.$

The sum of the integers between $1$ and $100$ inclusive that are divisible by $5$ is

$\displaystyle\sum_{k=0}^{20} (5k) = 5*\dfrac{20*21}{2} = 1050.$

The sum of the integers between $1$ and $100$ inclusive that are divisible by $15$ is

$\displaystyle\sum_{k=0}^{6} (15k) = 15*\dfrac{6*7}{2} = 315.$

By the principle of inclusion/exclusion, the sum of the integers between $1$ and $100$ inclusive which are divisible by $3$ or $5$ is then $1683 + 1050 - 315 = 2418.$

The sum of the integers between $1$ and $100$ inclusive which are not divisible by $3$ or $5$ is then

$\dfrac{100*101}{2} - 2418 = \boxed{2632}.$

Sum of all numbers 1 to 100: S = n[2a1 + (n - 1)d]/2 = 100[2(1) + (100 - 1)]/2 = 100(101)/2 = 5050

Sum of all numbers 3 to 99, multiples of 3: S = n[2a1 + (n - 1)d]/2 = 33[2(3) + (33 - 1)3]/2 = 33(6 + 96)/2 = 1683

Sum of all numbers 5 to 100, multiples of 5: S = n[2a1 + (n - 1)d]/2 = 20[2(5) + (20 - 1)5]/2 = 20(10 + 95)/2 = 1050

Sum of all numbers 15 to 90, multiples of 15: S = n[2a1 + (n - 1)d]/2 = 6[2(15) + (6 - 1)15]/2 = 6(30 + 75)/2 = 315

Sum of integers from 1 to 100 which are not divisible by 3 and 5: S = sum(1-100) - sum(3-99) - sum(5-100) + sum(15-90) = 5050 - 1683 - 1050 + 315 = 2632

Required sum = sum of numbers from 1 to 100 – sum of numbers from 1 to 100 which are divisible by 3 – sum of numbers from 1 to 100 which are divisible by 5 + sum of numbers from 1 to 100 which are divisible by 15 (LCM of 3 and 5)

Required sum = [1+2+3...+100] –[3+6+9+ ...+ 99] – [5+10+15+ ...+100] + [15+30+45+...+90]

= [1+2+3...+100] –[3(1+2+3+ ...+ 33)] – [5(1+2+3+ ...+ 20)] + [15(1+2+3+...+6)]

= [100×101/2] –[3×33×34/2)] – [5×20×21/2)] + [15×6×7/2)]

= 5050 – 1683 – 1050 + 315 = 2418

As in sum of numbers divisible by 3 and sum of numbers divisible by 5 contains common numbers which are divisible by 15 (15, 30, 45,..., 90) get subtracted so this sum is added.