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Re: Number with n^3 , 109
Posted:
Jan 2, 2013 11:37 AM
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On Jan 1, 9:52 am, "mina_world" <mina_wo...@hanmail.net> wrote:
> Hello, teacher~
>
> n^3 + (n+1)^3 + (n+2)^3 = 57 + 59 + 61 + ... + 109
>
> Find the "n".
>
> ------------------------------------------------------------------------
> Answer is 8.
>
> http://board-2.blueweb.co.kr/user/math565/data/math/n3.jpg
>
> This is a solution of elementary schoolchild(of course, clever)
>
> Can you understand it ? I need your explanation.(used formula etc)
Given n^3 +(n+1)^3+(n+2)^3=57+59+61+...+105+107+109
Find n
Interpretation of your student's solution:
==========================
The student solution uses 2 formulas and both of them are based on
mathematical induction.
(1) 1^3 + 2^3 +3^3 +...+n^3 = [n(n+1)/2]^2
(2) 1+2+3+...+n = n(n+1)/2
Use of (2) in the solution given by your student is a little tricky
but correct.
Part I: Finding sum of alternate odd numbers from 1 to 109 and from 1
to 55
=================================================
in the form on RHS of (1)
thus,
(3) sum 1 to 109 for odds = (109+1)*(55/2) = 55^2 (there are 55/2
pairs of numbers each pair with identical sum of 110)
(4) sum 1 to 55 for odds = (55+1)*(28/2) = 28^2 (same logic as
above in (3))
Part II: Using (1)
==============
Therefore, from given problem
n^3 +(n+1)^3+(n+2)^3= 55^2 -27^2
=> n^3 +(n+1)^3+(n+2)^3= (1+2+3+...+10)^2 -(1+2+...+7)^2
=> n^3 +(n+1)^3+(n+2)^3= [10(10+1)/2]^2 -[7(7+1)/2]^2 which now can
use (1) as below
=>n^3 +(n+1)^3+(n+2)^3= [1^3+2^3+...+10^3] - [1^3+2^3+...+7^3]
=>n^3 +(n+1)^3+(n+2)^3= 8^3 +9^3 +10^3
=> n=8 Ans.
NB: Pl. ignore typos if any
Mohan Pawar
Online Instructor, Maths/Physics
MP Classes LLC
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US Central Time: 10:37 AM 1/2/2013
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