+0  
 
0
63
4
avatar

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 \[a + (a + 1) + (a + 2) + \dots + (a + n - 1) = 100.\]

Guest Aug 26, 2018
 #2
avatar
0

 Sorry but I was hoping for a better explanation and solution because I didn't understand it...

Guest Aug 26, 2018
 #3
avatar+7023 
+1

Part(a):

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

MaxWong  Aug 27, 2018
 #4
avatar
0

What would part b be?? Thank you so much!!!

Guest Aug 27, 2018

10 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.