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Noticing Patterns In Addition

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Mathematics
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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. Each 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) x 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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Surname 1 Name Tutor Course Date 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. Each 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) x n. So, 2s= (n+1) n. Divide bo ...
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