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How would I sum the integers from 1 to 100, or from 1 to n?This is a question for my...

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crazykitty95 | Student, Grade 9 | eNotes Newbie

Posted August 18, 2009 at 11:47 AM via web

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How would I sum the integers from 1 to 100, or from 1 to n?

This is a question for my math homework and this is the only one I have having isues with I am completely stuck and have no idea how to do this pleeze help!

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neela | High School Teacher | (Level 3) Valedictorian

Posted August 18, 2009 at 12:46 PM (Answer #1)

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To calculate the sum, 1+2+3+4+......+98+99+100.

Let S be the sum of integers from 1 to 100.

Then,

S=1+    2+ 3+  4+ 5+.......+98+99+100. ---(1). Also if you reverse right side you get:

S=100+99+98+97+96....    + 3+ 2+   1-----(2)

Both (1) and (2) have 100 terms each on right side.

Add (1) and (2) vertically, term by term on tright side, we get:

2s=101+101+101+101+101+......101+101+101, There are 100 terms each being 101

Threfore,

2s=101*100

Divide both sides by 2 we get the required sum s:

2s/2 =101*100/2=

s= 101*50=5050.

 

Part (2): To calculate 1+2+3+4+5+.....(n-4)+.(n-3)+(n-2)+(n-1)+n.

Let s= 1+2+3+4+5+.......(n-2)+(n-1)+n           (1)

There are n terms. Rach term increase by 1 from the previous term. This is an arithmetic progression with a starting term 1 and common difference 1 and the number of terms  being n.

Reverse the right side of (1) as below:

s= n+(n-1)+(n-2)+(n-3)+(n-4)+...3+2+1           (2)

Add (1) and (2), particularly the right side vertically term by term:

2s= (n+1)+(n+1)+(n+1)+(n+1)+(n+1).......(n+1)+(n+1)+(n+1) .   There are n  terms like (n+1), whose sum is (n+1)*n. So,

2s=(n+1)n.

Divide both sides by 2 to get

2s/2=(n+1)n/2  or

s=n(n+1)/2. Therefore,

1+2+3+.....+n = n(n+1)/2

 

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krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted August 18, 2009 at 11:57 AM (Answer #2)

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There is a simple formula for calculating sum of integers from 1 to any given number.

Sum of integers from 1 to n = [n*(n + 1)]/2

Using this formula sum of integers from 1 to 100 = [100*(100 + 1)]/2

= (100*101)/2 = 5050

You can see how this formula works by taking a simple example of sum of integers from 1 to 10.

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 +10 can be rearranged as

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

This is = 11 + 11+ 11+ 11+ 11 = 11*5 = (10*11)/5

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rsarvar1a | Student, Grade 9 | TA | (Level 1) Honors

Posted April 19, 2015 at 7:02 PM (Answer #3)

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To calculate this sum, we need to realize that in the range of numbers from 1 to 100, there are a number of pairs that add up to 101. The number of pairs is half the number of integers, that is to say, 50 pairs. This means, in effect, that in the range of numbers from 1 to 100, there are 50 numerical pairs of 101.

We can express this as an equation:

[n * (n + 1)] / 2 = sum of 1 to n.

To plug in the range of numbers from 1 to 100, we do this:

[100 * (100 + 1)] / 2 = sum

(100 * 101) / 2 = sum

10100 / 2 = sum

5050 = sum

Therefore, the sum of numbers 1 to 100 is 5050.

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