the following two lines indicate another way to derive the formula for the sum of the first n integers by rearranging the terms in the sum. Fill in the details.
1+2+3+...+n=(1+n)+(2+(n-1))+(3+(n-2))+...
=(1+n)+(1+n)+(1+n)+...
...
= (1+n)+(1+n)+(1+n)+... for a total of n/2 such binomials
= (n/2)(n+1
or
n(n+1)/2 or (1/2)(n)(n+1) or (n^2 + n)/2
To derive the formula for the sum of the first n integers by rearranging the terms in the sum, let's break down the process step by step:
We start with the sum of the first n integers:
1 + 2 + 3 + ... + n
Now, we can rearrange the terms in the sum by pairing the first term with the last term, the second term with the second-to-last term, and so on:
(1 + n) + (2 + (n - 1)) + (3 + (n - 2)) + ...
Let's evaluate the first few terms to see the pattern:
(1 + n) = (n + 1)
(2 + (n - 1)) = (n + 1)
(3 + (n - 2)) = (n + 1)
As you can see, each pair of terms adds up to (n + 1). We can continue this pattern until we reach the middle term, which will also be paired with itself:
(n + 1) + (n + 1) + (n + 1) + ... + (n + 1)
We are essentially adding (n + 1) a total of n times. So, the simplified expression becomes:
(n + 1) * n
Therefore, the formula for the sum of the first n integers derived by rearranging the terms in the sum is:
1 + 2 + 3 + ... + n = (n + 1) * n