April 5th, 2014, 09:27 AM  #1 
Newbie Joined: Apr 2014 From: Earth Posts: 14 Thanks: 0  How to solve this consecutive numbers addition?
Hello everyone! Please teach me how to solve this question: Find the value of: 7+9+11+13+15+17+……+55+57+59+61 Ans: 952 and this question: Find the value of: 10+20+3+4+5+……+6+7+8+9+10 Thanks! 
April 5th, 2014, 10:07 AM  #2 
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 the first one, we could use the formula: $\displaystyle \sum_{k=1}^n(2k1)=n^2$ And so your sum $S$ is: $\displaystyle S=\sum_{k=4}^{31}(2k1)=\sum_{k=1}^{31}(2k1)\sum_{k=1}^{3}(2k1)=31^23^2=(31+3)(313)=28\cdot34=952$ 
April 5th, 2014, 10:26 AM  #3 
Newbie Joined: Apr 2014 From: Earth Posts: 14 Thanks: 0 
Hello MarkFL, Thanks for sharing! Can you share me the Primary Schools method too? Thanks, XP 
April 5th, 2014, 10:39 AM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 
Well, we could state: $\displaystyle S=7+9+11+\cdots+57+59+61$ $\displaystyle S=61+59+57+\cdots+11+9+7$ Now, if we add each column, we find: $\displaystyle 2S=68+68+68+\cdots+68+68+68$ Seeing that we have 28 68's on the right side, we may then write: $\displaystyle 2S=28\cdot68$ Divide both sides by 2: $\displaystyle S=28\cdot34=952$ 
April 5th, 2014, 11:32 AM  #5 
Newbie Joined: Apr 2014 From: Earth Posts: 14 Thanks: 0 
Hi MarkFL, The 2nd method looks easier but I don't understand. 1. What is "S"? 2. What is "2S"? 3. How did you know there are 28 68's on the right without calculating? 4. Final step, S= 28.34, how did you calculate into 952? 5. Add what column? Do you still have any other easier and simpler steps? Regards, Mai 
April 5th, 2014, 11:43 AM  #6 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 
$S$ just stands for the value of the sum, and so $2S$ is twice the value of the sum. $7=2\cdot41$ and $61=2\cdot311$, and so there are $314+1=28$ addends in the sum. $\displaystyle 28\cdot34=4\cdot14\cdot17=4(151)(15+2)=4(225+152)=4(238)=4(2402)=9608=952$ If you look at the two expressions I wrote, then look from top to bottom at each term that lines up: $\displaystyle S+S=(7+61)+(9+59)+(11+57)+\cdots+(57+11)+(59+9)+(6 1+7)$ $\displaystyle 2S=68+68+68+\cdots+68+68+68$ Last edited by MarkFL; April 5th, 2014 at 11:47 AM. 
April 5th, 2014, 09:15 PM  #7  
Newbie Joined: Apr 2014 From: Earth Posts: 14 Thanks: 0 
Hello MarkFL, Then what about the dot (.) stand for? Quote:
Best regards, Mai  
April 5th, 2014, 10:01 PM  #8 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 
That dot is above where a decimal point would go and is a symbol for multiplication. 
April 5th, 2014, 10:55 PM  #9  
Newbie Joined: Apr 2014 From: Earth Posts: 14 Thanks: 0 
Hi MarkFL, Do you have any other easier methods? Your 3rd method is more confusing. Quote:
Quote:
By the way, the 2nd method I understood a little bit, I understood how you got 68 but not understand how you got 28 68's. Quote:
Mai Last edited by XPMai; April 5th, 2014 at 11:04 PM.  
April 5th, 2014, 11:20 PM  #10 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 
Well, you could manually add the terms. The 4 and 31 come from: $\displaystyle 7=2\cdot41$ $\displaystyle 61=2\cdot311$ The 1 comes from the fact that the number of terms is the difference in indices plus 1.For example, consider if you have 10 objects labeled 1 through 10. (10  1) + 1 = 10 

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