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Topic:
Number with n^3 , 109
Replies:
5
Last Post:
Jan 10, 2013 5:42 AM




Re: Number with n^3 , 109
Posted:
Jan 2, 2013 11:37 AM


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://board2.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  US Central Time: 10:37 AM 1/2/2013



