Question

sum of all even integers between 99 and 303

Answer 1

I must say... people who make this sort of question are pure evil... or think their students have nothing better to do with their time :D

Answer 2

So, the series is 100 + 102 + 104 + ... + 300 + 302

Answer 3

Notice that the first term and the last term add up to 402

Answer 4

The second term and the second-last term add up also to 402

Answer 5

So would the third term and the third-last term

Answer 6

There are 102 terms, so there would be 61 pairs which add up to 402

Answer 7

So the sum is 61 x 402 = 24522

Answer 8

@kc_kennylau You definitely got the right idea here, but I think you miscounted the number of terms between 99 and 303.

There are \(303-99 + 1 = 205\) terms between. Since 205 is an odd number, we can't find "pairs" for every number, so we'll ignore the 205th number which is 303.
Thus there are 204 terms between 99 and 302, thus giving us 102 pairs.

So, you have 99, 100, .., 301, 302. THe first and last term's sum is 99+302 = 401
Then the sum of 100 + 301 = 401, etc... there are 102 pairs like this;
\(102 \times 401 = 40902\). Now, we are just missing to add the last term "303". So \(40902+303 = 41205\)

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Another method is to use the formula for the sum of integers.
\[ \large \sum_{i=99}^{303}i=\sum_{i=1}^{303}i-\sum_{i=1}^{98}i=\frac{303(304)}{2}-\frac{98(99)}{2}=41205\]

Answer 9

@kirbykirby You misread the question. It asks for the sum of all \(\fbox{even}\) integers in [99,303].

Answer 10

oh dear @_@ how embarrassing. I totally missed that, thanks for pointing it out.