My Math Forum [ASK] Induction Again

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 February 2nd, 2018, 05:54 PM #1 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 1,993 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: 428 Thanks: 144 Hint:
 February 2nd, 2018, 09:52 PM #3 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 1,993 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: 428 Thanks: 144 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: 1,993 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: 19,174 Thanks: 1644 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: 1,993 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: 19,174 Thanks: 1644 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 $S_1 + S_2 + S_3 +\ ... +\ S_{n-1} = 364$. 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. Thanks from Monox D. I-Fly
 February 4th, 2018, 08:28 PM #9 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 1,993 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: 7,822 Thanks: 1049 Math Focus: Elementary mathematics and beyond Using $\displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{ 6}$ and $\displaystyle\sum_{k=1}^nk=\frac{n(n+1)}{2}$ $$\frac12\sum_{k=1}^nk^2-k=\frac{2n^3+3n^2+n}{12}-\frac{n^2+n}{4}=\frac{n^3-n}{6}=364\implies n^3-n=2184\implies n>10$$ 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 We see that the differences are constant on the 2nd line down so we seek a polynomial of degree 2, in the form $\displaystyle an^2+bn+c$. This gives rise to a system of equations: $$a+b+c=1 \\ 4a+2b+c=3 \\ 9a+3b+c=6$$ $$3a+b=2 \\ 8a+2b=5$$ $$a=b=\frac12,\,c=0$$ 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. Thanks from Monox D. I-Fly

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