Question

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

Answer 1

Answer:

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

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

Step-by-step 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 $\frac{101 \cdot (202)}{2}= 101^2 = 10201$

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