Fasrt Math Help? It's easy? 4th Grade..?

Question

3. the sum of the first 10 counting numbers (1 point)

4. the sum of the first 1,000 counting numbers

Answer 1

The sum of the first n numbers {1, 2, ...n} is n*(n+1)/2

So. the sum of the first 10 counting numbers is 10*11/2 = 55

The sum of the first 1000 counting numbers is 1000*1001/2 = 500500

I am assuming you are in 4th grade, so I will explain where this equation comes from.

Let S = the sum of the first n counting numbers

S = 1 +   2  + 3 + ... + n

S = n + n-1 + ....      + 1

Then, sum up the two above formulas to get 2S = n(n - 1)

Notice each term adds up to n-1 and there are n of them.

So, to solve for S, we divide by 2 on either side and get S = n*(n-1)/2.

Good luck.

Answer 2

you have to find 1+9(10) then, 10 divided by 2 is 5. so 5 times 10 is 50. the answer is 50.(logic is 1+9=10, 2+8=10etc) same 4 4..

Answer 3

if you don't want to figure out something this stupid what makes you think other people will

Answer 4

what are the first 10 counting numbers? 1, 2, 3, 4.....9, 10. so what is the sum of them? Add them up. 1+2+3+4+.....+9+10

Answer 5

you can use a calculator to add it... but it will take so much time when you do that especially in number 4

there is a way you can do it mentally

there is a addition pattern in finding the sum of consecutive positive integers

multiply the largest number(n) by the "largest number +1" (n+1) and divide the product by 2.

sum of counting numbers from 1 to n = n(n+1) / 2

#3 the sum of the first 10 counting numbers

the largest number is 10

then from the equation n(n+1) / 2

10 (10+1)/2

10 (11) / 2

110/2

=55

the sum of 1 to 10 is 55

OK now lets try the number 4

the sum of 1 to 1000

the largest number is 1000

apply it to the formula I gave you

1000 (1000+1)/2

1000(1001) /2

1001000 / 2

=500500

the sum of 1 to 1000 is 500500

its so easy... just remember the formula

hope it helps...GUDLUCK!!! ^^,

Answer 6

for question 4. use the equation

[n(n+1)]/2