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Q. re number propeties

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Q. re number propeties

by gmat1011 » Thu Jul 01, 2010 7:41 am
The MGMAT book provides a useful maths rule:

- sum of an odd number (total of N) of consecutive integers is always divisible by N
- such sum is never divisible by N if N is even

Does any one know of any similar rule when it comes to a sum of an evenly-spaced series of numbers, such as say the first 10 even numbers or the first 5 odd numbers

Basically want to know if the number property in question only applies when the series is evenly spaced in increments of 1 only (i.e., consecutive integers) or does it apply to any evenly spaced set of numbers pr consecutive multiples?

Thanks very much.
Last edited by gmat1011 on Thu Jul 01, 2010 8:54 am, edited 1 time in total.
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by kvcpk » Thu Jul 01, 2010 8:26 am
gmat1011 wrote:The MGMAT book provides a useful maths rule:

- sum of an odd number (total of N) of consecutive integers is always divisible by N
- such sum is never divisible by N if N is even

Does any one know of any similar rule when it comes to a sum of a evenly-spaced series of numbers, such as say the first 10 even numbers or the first 5 odd numbers

Basically want to know if the number property in question only applies when the series is evenly spaced in increments of 1 only (i.e., consecutive integers) or does it apply to any evenly spaced set of numbers pr consecutive multiples?

Thanks very much.
Thanks for sharing the rule.. I didnt know this..

Coming to your query.. I can give you examples for which it doesnt work for equally spaced integers. (for the second rule)
let the numbers be
-2,0,2,4
Nis even = 4
Sum = 4
4 is divisible by 4. So 2nd rule is violated.
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by albatross86 » Thu Jul 01, 2010 9:11 am
The sum of the first n integers = n*(n+1) / 2

The sum of the first n odd numbers = n^2

The sum of the first n even numbers = n*(n+1)
~Abhay

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by gmat1011 » Thu Jul 01, 2010 9:43 am
Thanks for sharing

Do these addition formulae work if you say want to consider the sum of even integers between 50 and 150 (101 integers)... that is if the sequence is not starting form the lowest possible integer in that sequence.... thx.
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by albatross86 » Thu Jul 01, 2010 10:03 am
All these formulas only apply to the "first n" of these sequences.

The easiest way to solve a generic sequence, in which the difference between two consecutive terms is constant, is to apply the Arithmetic Progression formulae. These are:

A[n] = A[1] + (n - 1) * d ..... Eqn 1

S[n] = n/2 * { 2*A[1] + (n - 1)*d } ..... Eqn 2

Where A[n] = nth term in the sequence, A[1] = 1st term in the sequence. d = constant difference. S[n] = sum of n terms.

You can use these to derive those formula I mentioned above, as well as to solve any sequence thrown at you. It's useful when a graceful / conceptual approach isn't evident.

eg. What is the sum of all the odd numbers from 54 to 88?

Solution: The first odd number, i.e. A[1] = 55. The last odd number, i.e. A[n] = 87. The difference between any two consecutive odd or even integers is d = 2

Thus from Eqn 1:

87 = 55 + (n - 1) * 2

=> solve for n to get n = 17

Now use Eqn 2:

S[n] = 17/2 * {2*55 + (17 - 1)*2}
= 17/2 * {110 + 32}
= 17/2 * 142
= 17 * 71

S[n] = 1207 ... Ans


There is another way to look at this too. If you consider the arithmetic mean or average of a set of odd integers, it is always equal to the median, which is itself equal to ( A[1] + A[n] ) / 2 .....In this case this equals 71

We know that Arithmetic mean = Sum of terms / Total number of terms

=> 71 = Sum / 17
=> Sum = 1207


These different approaches to problems will all come in handy in different situations.
~Abhay

Believe those who are seeking the truth. Doubt those who find it. -- Andre Gide
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by selango » Thu Jul 01, 2010 9:32 pm
Let me add few shortcuts,

Natural number=Positive integers

Sum of squares of first n natural numbers,

S=1^2+2^2+3^2+....n^2

S={n(n+1)(2n+1)/6}

-------------------------------------------------------------

Sum of cubes of first n natural numbers,

S=1^3+2^3+3^3+....n^3

S=[n(n+1)/2]^2

--------------------------------------------

Sum of odd numbers <=n,where n is the natural number

n is odd-- [(n+1)/2]^2

n is even--[n/2]^2


--------------------------------------------------------------------------

Sum of even numbers <=n,where n is the natural number

n is odd-- [(n-1)/2*(n+1)/2]

n is even--[n/2*(n/2+1)]
--Anand--
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