February 2nd, 2018, 05:54 PM  #1 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 2,000 Thanks: 132 Math Focus: Trigonometry  [ASK] Induction Again
Determine the value of n if 1 + 3 + 6 + ... + $\displaystyle \frac{1}{2}$n(n  1) = 364 What I did: I know that 1, 3, and 6 are the result of arithmetic series with the starting value 1 and the difference 2, thus that sum can be written as $\displaystyle S_1$ + $\displaystyle S_2$ + $\displaystyle S_3$ + ... + $\displaystyle S_n$ = 364. However, by assuming that $\displaystyle S_n$ = $\displaystyle \frac{1}{2}$n(n  1) I got n + 1 = n  1 which is simply unsolvable at all. After all, the term $\displaystyle \frac{1}{2}$n(n  1) doesn't match for n = 1. Does this question even have any solution? 
February 2nd, 2018, 09:13 PM  #2 
Senior Member Joined: Oct 2009 Posts: 494 Thanks: 164 
Hint: 
February 2nd, 2018, 09:52 PM  #3 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 2,000 Thanks: 132 Math Focus: Trigonometry 
I knew them already, but should I manually put the sums one by one in order to get n?

February 2nd, 2018, 10:26 PM  #4 
Senior Member Joined: Oct 2009 Posts: 494 Thanks: 164 
No, you should find a formula by induction. For example: $$1+2+3+...+n = \frac{n(n+1)}{2}$$ 
February 2nd, 2018, 11:08 PM  #5 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 2,000 Thanks: 132 Math Focus: Trigonometry 
After that, how to determine the sum of that formula?

February 3rd, 2018, 11:29 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,532 Thanks: 1750 
Each $S_n$ is the sum of the first $n$ natural numbers. Let $U_n$ denote the sum of the first $n$ of the above sums. Hint: calculate $U_n/S_n$ for successive values of $n$. 
February 4th, 2018, 05:23 PM  #7 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 2,000 Thanks: 132 Math Focus: Trigonometry 
I still can't comprehend it. Can you explain them step by step?

February 4th, 2018, 06:39 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 19,532 Thanks: 1750 
The natural numbers form an A.P. (arithmetic progression). The sum of $n$ consecutive terms of an A.P. is $n$ times the mean of the first and last terms in the sum. Hence $S_n = n(1 + n)/2$. The question asks for the value of $n$ such that $S_1 + S_2 + S_3 +\ ... +\ S_{n1} = 364$. One can equivalently solve $U_n = S_1 + S_2 + S_3 +\ ... +\ S_n = 364$ for $n$ and then add 1 to the answer. Calculating $S_1$, $S_2$, $S_3$, $S_4$, $S_5$, $S_6$, ... gives 1, 3, 6, 10, 15, 21, ... and so the corresponding values of $U_n$ are 1, 4, 10, 20, 35, 56, ... and the corresponding values of $U_n/S_n$ are 1, 4/3, 5/3, 2, 7/3, 8/3, ... so it's apparent that $U_n/S_n = (n + 2)/3$, which implies $U_n = ((n + 2)/3)(n(1 + n)/2) = n(n + 1)(n + 2)/6$. Hence one needs to solve $n(n + 1)(n + 2) = 364\times6 = 4\times7\times13\times6 = 12\times13\times14$. That has solution $n = 12$. Hence the solution to the original problem is 13. 
February 4th, 2018, 08:28 PM  #9 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 2,000 Thanks: 132 Math Focus: Trigonometry 
Thank you, Skip! I can follow this one, though I wonder if I can explain it well to my partner.

February 5th, 2018, 03:33 AM  #10 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,859 Thanks: 1080 Math Focus: Elementary mathematics and beyond 
Using $\displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{ 6}$ and $\displaystyle\sum_{k=1}^nk=\frac{n(n+1)}{2}$ $$\frac12\sum_{k=1}^nk^2k=\frac{2n^3+3n^2+n}{12}\frac{n^2+n}{4}=\frac{n^3n}{6}=364\implies n^3n=2184\implies n>10$$ To derive $\displaystyle\sum_{k=1}^nk=\frac{n(n+1)}{2}$, write out the first four terms of the sum then take successive differences: Code: 1 3 6 10 2 3 4 1 1 $$a+b+c=1 \\ 4a+2b+c=3 \\ 9a+3b+c=6$$ $$3a+b=2 \\ 8a+2b=5$$ $$a=b=\frac12,\,c=0$$ so our polynomial is $\displaystyle\frac{n^2+n}{2}=\frac{n(n+1)}{2}$. Proof by induction: Base case: $$n=1\Rightarrow\frac{1(1+1)}{2}=1$$ Inductive step: $$ \frac{(n+1)(n+2)}{2}=\frac{n^2+2n+n+2}{2}=\frac{n( n+1)}{2}+(n+1)$$  One could do the same for the cubic but I won't post that here. 

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