Question

7. Find the following sums: a. 2 4+6 +8+ 10.. +200 b. 51+52 + 53+54+... +151

Answer 1

The sum of
`2+4+6+8+10+... +200 = 10100`

The sum of
`51+52+53+54+...+151 = 10201`

Explanation:

(a) To find the sum of
`2+4+6+8+10+... +200`, take the first and the last number 2 + 200 = 202.

Now that those are accounted for, take the next smallest and largest numbers available: 4 + 198 = 202.

Continuing, 6 + 196 = 202

This repeats 50 times until reaching 50 + 152 = 202.

Since 202 repeats 50 times, the desired sum is 50 x 202 = 10100

Therefore
`2+4+6+8+10+... +200 = 10100`

To check the result we can use the formula to sum of the first n even numbers:

`n \cdot (n+1)`

where n in our case is 100 because

`2+4+6+8+10+... +2\cdot(100)`

so
`100 \cdot (100+1)=10100`

(b) To find the sum of
`51+52+53+54+...+151`, do the same in point (a)

51 + 151 = 202

52 + 150 = 202

53 + 149 = 202

Note that there are (151 - 51 + 1 ) = 101 terms

so the sum will be
`(101 \cdot (202))/(2)= 101^2 = 10201`

Therefore
`51+52+53+54+...+151 = 10201`