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Part (a): Find the sum \[a + (a + 1) + (a + 2) + \dots + (a + n - 1)\] in terms of  $a$ and  $n$

Part (b): Find all pairs of positive integers $(a,n)$ such that $n \ge 2$ and \[a + (a + 1) + (a + 2) + \dots + (a + n - 1) = 100.\]

 Aug 26, 2018
 #2
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 Sorry but I was hoping for a better explanation and solution because I didn't understand it...

 Aug 26, 2018
 #3
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Part(a):

\(\quad a+(a+1)+(a+2)+\dots+(a+n-1)\\ =(1+2+3+4+5+...+(a+n-1)) - (1+2+3+4+5+...+a-1)\\ \boxed{\text{Know: }1+2+3+4+...+ n=\dfrac{n(n+1)}{2}}\\ =\dfrac{(a+n-1)((a+n-1)+1)}{2} - \dfrac{(a-1)((a-1)+1)}{2}\\ =\dfrac{(a+n)(a+n-1)}{2}-\dfrac{a(a-1)}{2}\\ =\dfrac{a^2+2an+n^2-a-n}{2}-\dfrac{a^2-a}{2}\\ =\dfrac{2an+n^2-n}{2}\\ =an+\dfrac{n(n-1)}{2}\)

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 Aug 27, 2018
 #4
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What would part b be?? Thank you so much!!!

Guest Aug 27, 2018

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