April 6th, 2014, 12:14 AM  #11  
Newbie Joined: Apr 2014 From: Earth Posts: 14 Thanks: 0 
Hello Mark FL, What if the smallest number is an even number and the largest number also even number, then how to multiply and minus 1? Example this question mentioned at 1st: Quote:
Quote:
Quote:
$\displaystyle 10✕1=10$ But answer seems to be wrong... Kindest regards, Mai  
April 6th, 2014, 12:24 AM  #12 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 
For a sum of the first $n$ natural numbers, we can use the formula: $\displaystyle \sum_{k=1}^nk=\frac{n(n+1)}{2}$ In this case, $n=10$ and so we have: $\displaystyle \sum_{k=1}^{10}k=\frac{10(10+1)}{2}=55$ 
April 6th, 2014, 12:38 AM  #13 
Newbie Joined: Apr 2014 From: Earth Posts: 14 Thanks: 0 
Hi MarkFL, This method is too difficult for me, can we use the easier method? The state method. High regards, Mai 
April 6th, 2014, 01:19 AM  #14 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 
What is the state method? If you are going to do sums, then you need to know these formulas eventually. edit: You could state: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = (1 + 10) + (2 + 9) + (3 + 8) + (4 + 7) + (5 + 6) = 5(11) = 55 But this would be tedious if you had the first 1000 natural numbers. Last edited by MarkFL; April 6th, 2014 at 01:27 AM. 
April 6th, 2014, 03:24 AM  #15  
Newbie Joined: Apr 2014 From: Earth Posts: 14 Thanks: 0 
Hello MarkFL, This is the state method you said: Quote:
Highest regards, Mai  
April 6th, 2014, 03:27 AM  #16 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 
Yes, try it...write the sum both ways, then add them up.

April 6th, 2014, 09:50 AM  #17  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,414 Thanks: 1024  Mark showed you the "tricks of the trade" used to arrive at the formula(s). This series is a bit different from the usual; given is the 1st term (7), last term (61) and common difference (2). The number of terms n (28 in your example) is not given; usually it is... a = 1st term (7) b = last term (61) d = common difference (2) n = ? n = (b  a + d) / d : that's the formula to calculate n so: n = (61  7 + 2) / 2 = 56 / 2 = 28 The standard formula for the sum is : n(a + b) / 2 So we can calculate the sum of your series this way: S = n(a + b) / 2 where n = (b  a + d) / d S = 28(7 + 61) / 2 = 952 And don't ask for a simpler way...cause there ain't none! Perhaps reading the stuff here will help you: Arithmetic Series Quote:
n = (10  1 + 1) / 1 = 10 (of course!) S = 10(1 + 10)) / 2 = 55 However, this one is a "special" case: sum of first n positive numbers. Since a is always 1 and n is always b, then substituting in above formula: S = n(1 + n) / 2 S = 10(1 + 10) / 2 = 55 And this explains the standard formula S = n(n + 1) / 2 for the sum of the first n positive numbers. OK? Last edited by Denis; April 6th, 2014 at 09:53 AM.  
April 7th, 2014, 03:58 AM  #18  
Newbie Joined: Apr 2014 From: Earth Posts: 14 Thanks: 0 
Hehe, I've finally found out the method that I was looking for! Question: Quote:
Quote:
So that is the easiest method as compared to you guys. Cheers, Mai Last edited by XPMai; April 7th, 2014 at 04:23 AM. Reason: Addon: So that is the easiest method as compared to you guys.  
April 7th, 2014, 05:02 AM  #19 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,414 Thanks: 1024 
You find that easier than 10*11 / 2 = 55 ?

April 7th, 2014, 05:36 AM  #20  
Newbie Joined: Apr 2014 From: Earth Posts: 14 Thanks: 0  Quote:
Yes, I found it easier to solve the question using my method than your methods or MarkFL's method because I can't understand your methods and MarkFL's methods. You can use my method in future too because it's easiest. Best wishes, Mai  

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