1+2+3+4+.......+999+1000 computing method please?

Question

what is the easy method for primary student to compute this calculation

Answer 1

Too add all the integers between 1 and n, the formula is:

[n(n + 1)]/2

In this case, n is 1000

so [1000*1001]/2

or 500500

Answer 2

Use a for .... next loop running from 1 to 1000, adding the value of the loop into another variable with each pass. When you get to the end the variable will have the total of 1+2+3...+999+1000. You can then show the user the value of the variable.

In VB, do this

dim intTotal as Integer
dim intCounter as Integer

intTotal = 0

for intCounter = 1 to 1000
intTotal = intTotal + intCounter
next intCounter

msgbox (intCounter)

Answer 3

One way to do it simply is this...
Notice that 1+999 = 1000, 2+998=1000, 3+997=1000, etc, all the way to 499+501=1000

So, you have 499 combinations listed there. The only numbers from the original list you didn't use in those combinations are 500 and 1000.

499(1000) + 500 + 1000
499000 + 1500
500500

Answer 4

1+2+3+......+n = m

Hopefully can help
easy to remember

even example n=10
1. Take the middle value = 5
2. Add with the next number = 6 so 5+6 = 11
3. Times by middle value =5 so 11*5 = 55

odd example n=5
1. Assume n is even value below 5 = 4
2. Take the middle value = 2
3. Add with the next number = 3 so 2+3= 5
4. Times by middle value =2 so 5*2=10
5. Add the latest number n itself = 5 so 10+5 = 15

actually step 1,2,3 for even is same 2,3,4 for odd

If want to using formula use formula below :


even formula
m = (n^2 + n) /2

ex:
n = 10
m =(10^2 +10) /2
m=110/2
m=55

odd formula
m = ((n^2-n) /2 )+n

ex:
n=5
m = ((5^2-5) /2 )+5
m = (25-5)/2 + 5
m = 20/2 + 5
m = 10 + 5 =15

Answer 5

teach them the pattern that occurs when adding 1~10...

1+2+3+4+5+6+7+8+9+10

teach them that adding...
1+10 = 11
2+9 = 11
3+8 = 11
and so on... then ask them how many sets of 11's there are.
since there are 5 sets of 11's... the sum of 1~10 is 5*11=55.

Now ask the student what pattern he recognizes and try applying it to the sum of 1~100.

When the student realizes the pattern of 50 set of 101, and gets 5050 as the answer, then he/she will be able to do 1~1000.

Just be patient.

Answer 6

Write the numbers from 1 to 500, then underneath them write the numbers from 1000 down to 501.

Add up each pair: 1 + 1000 = 1001. 2 + 999 = 1001. Keep going.

Now add up all of these. Since they're all the same, all you need to know is how many of them there are.

What's 1001 + 1001 + 1001 + ... + 1001 (where there are 500 of them)?

Answer 7

A instructor, eager to maintain her scholars busy for a mutually as gave this comparable question to her type. It so happenened that certainly one of her scholars became Gauss, who grew to improve right into a properly-liked mathemetician To the academics astonisment youthful Gauss responded the question in basically a couple of minutes. His reasoning became: one thousand +one million =1001 999 +2 =1001 998 +3 =1001 and so forth Hes observed that bthere may well be 500 pairs of numbers that further to 1001 so he elevated 500* 1001 to get the respond 500,500. He had if truth be told created the formulation (n/2)(n+one million) the place n is nthe finished form of words. no longer undesirable for a yuung newborn, eh?

Answer 8

oh that's easy

tell them to use a calculator to add all numbers up.

but if you want them to learn not using any technology


tell them to write their soulution in a paper like this

1
+2
=3
+4
=7

and so on

teach them how to be patient

remember
"PAtience is a Virtue"

^_^

Answer 9

S = x(i+ l)/2
x= number of such numbers
S= Sum of numbers
i= first number
l= last number