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April 6th, 2014, 12:14 AM   #11
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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:
 Find the value of: $\displaystyle 1+2+3+4+5+...+9+10$
Quote:
 $\displaystyle 1=1✕1-0$
Quote:
 $\displaystyle 10=5✕2-0$
Final step:
$\displaystyle 10✕1=10$

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
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Hello MarkFL,

This is the state method you said:

Quote:
 $\displaystyle S=7+9+11+⋯+57+59+61$ $\displaystyle S=61+59+57+⋯+11+9+7$ Now, if we add each column, we find: $\displaystyle 2S=68+68+68+⋯+68+68+68$ Seeing that we have 28 68's on the right side, we may then write: $\displaystyle 2S=28⋅68$ Divide both sides by 2: $\displaystyle S=28⋅34=952$
So is it possible to use the same above method to solve the question?

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
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Quote:
 Originally Posted by XPMai Find the value of: 7+9+11+13+15+17+……+55+57+59+61
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!
Arithmetic Series
Quote:
 Find the value of: 1+2+3+4+5+6+7+8+9+10
You can use the same formula:
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
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Hehe, I've finally found out the method that I was looking for!

Question:
Quote:
 Find the value of: $\displaystyle 1+2+3+4+5+...+8+9+10$
Working:
Quote:
 Step 1: Pair the 1st and last number up and find the sum of them. ....$\displaystyle 1+2+3+4+5$ ...$\displaystyle 10+9+8+7+6$ $\displaystyle = 11....11...11..11...11$ (Ignore the dots, the dots are just to make them equal.) Step 2: Find the number of total pairs $\displaystyle 10÷2=5$ (pairs) Step 3: Multiple the number of pairs by the sum of 1 pair. $\displaystyle 5×11=55$

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: Add-on: 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: 14,597 Thanks: 1038 You find that easier than 10*11 / 2 = 55 ?
April 7th, 2014, 05:36 AM   #20
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Quote:
Originally Posted by Denis
Quote:
 Find the value of: 1+2+3+4+5+6+7+8+9+10
You can use the same formula:
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
Hello Denis,

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

### how to add consecutive nos using calculus

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