**1. Find the sum of even natural numbers 1 to 100.**

Ans- There are 50 even numbers between 1 and 100. Now, according to formulae, n(n+1) = 50×51= 2550.

Similarly, you can find the sum of odd natural numbers between 1 and 100.

**2. Find the sum of even numbers from 41 to 100.**

Ans- firstly you should find the sum of even numbers from 1 to 100 = 2550

Then, find the sum of even natural’s number from 1 to 40= 20×21= 420

Now, the sum of even numbers from 41 to 100= 2550-420= 2130

**3. Find the sum of the square of natural numbers 1 to 20.**

Ans- according to question, find 1

^{2}+2^{2}+3^{2}…. 20^{2}= n(n+1)(2n+1)/6Now put n=20,

= 20(21)(41)/6 = 2870

**4. Find the sum of the square of natural numbers from 11 to 20.**

Ans- according to question, find 11

^{2}+12^{2}+13^{2}….. 20^{2}= sum of the square from 1 to 20 – sum of the square from 1 to 10.Now, the sum of the square from 1 to 20 =2870

And, the sum of the square from 1 to 10 = 385

Hence, the sum of square from 11 to 20 = 2870-385= 2485.

**5. Find the sum of the square of natural numbers – 4**

^{2}+ 6^{2}+ 8^{2}+ 10^{2}+ 12^{2}+ 14^{2}+ 16^{2}+ 18^{2}+ 20^{2}.Ans- In these types of question, you can see

**is common so take out the 2***2*^{2}^{2}and then add the remaining values.Therefore, 2

^{2}(2^{2}+ 3^{2}+ 4^{2}+ 5^{2}+ 6^{2}+ 7^{2}+ 8^{2}+ 9^{2}+ 10^{2}) = 4×384 = 1536.**6. Find the sum of digits from 1 to 100.**

Ans- In these questions, one has to count each and every digit occurring from 1 to 100.

Now, from 1 to 100= ‘0’ occurs 11 times; ‘1’ occurs 21 times; and ‘2-9’ occurs 20 times each.

Hence, the sum of digits from 1 to 100 = (11×0) + (21×1) + (44×20) = 901.