Find the sum of the first 200 positive integers?

Answer 1

20,100

S200=2002(1+200)
S200=100(201)
S200=20100

Answer 2

Oh, dude, you want me to add up the first 200 positive integers? Like, seriously? Fine, it's just a little formula magic. The sum is n(n+1)/2, where n is the last number, so in this case, it's 200(200+1)/2. Do the math, and you get 20,100. Easy peasy, lemon squeezy.

Answer 3

This is easier than it looks. Imagine finding the sum of the first 8 integers. The obvious way is 1+2+3+4+5+6+7+8. If you slog through this you find the answer is 36. Now let's look at this differently. 1+8=9, 2+7=9, 3+6=9, 4+5=9. There is a pattern here ! All the integers got used, each pair adds up to 9, and there 4 pairs. Four nines make 36, which is the same answer as the other way. So, 1+200=201, 2+199=201, 3+198=201, . . . . 98+103=201, 99+102=201, 100+101=201. All the numbers got used, each pair summed to 201 and there were 100 pairs. 201 X 100 = 20100, and that is the sum of the first 200 integers. Saves a lot of wear and tear on the fingers !

Answer 4

This is easier than it looks. Imagine finding the sum of the first 8 integers. The obvious way is 1+2+3+4+5+6+7+8. If you slog through this you find the answer is 36. Now let's look at this differently. 1+8=9, 2+7=9, 3+6=9, 4+5=9. There is a pattern here ! All the integers got used, each pair adds up to 9, and there 4 pairs. Four nines make 36, which is the same answer as the other way. So, 1+200=201, 2+199=201, 3+198=201, . . . . 98+103=201, 99+102=201, 100+101=201. All the numbers got used, each pair summed to 201 and there were 100 pairs. 201 X 100 = 20100, and that is the sum of the first 200 integers. Saves a lot of wear and tear on the fingers !