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I am not sure how to properly do this question but I am told that the solution I came up with is wrong and I dont see how...I basically used algebra and plugging of variables and rearranging equations and used the given identities to prove my claim but its wrong so I hope someone can show me how to do this properly..

Above is the question and below is my solution. enter image description here

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What was the formula you came up with for $t_n$? –  JavaMan Jan 24 '12 at 6:19
    
You see the equation for Sn? I isolated Tn in there and then plugged that Tn into the equation for Tn so I had Tn= Sn-S(n-1) I have it circled or squared in my solution in middle of page..Also can you please tell me how you properly write subscripts and superscripts and formulas on this forum so in the future I can improve –  Raynos Jan 24 '12 at 6:23
    
When you were asked to find a formula for $t_n$, it is supposed to be independent of $s_n$. It should just be a formula $t_n = \text{some quantity in }n$. To see how to properly write a formula, you can right click on any equation you see and it will show the source code. In this case "$\$$t_n$\$$" produces $t_n$. –  JavaMan Jan 24 '12 at 6:27
    
$t_n=\frac{n(n+1)}{2}$ –  pedja Jan 24 '12 at 6:28
    
Your solution, to the degree I can understand it, assumes the result, makes a few substitutions, undoes them, and arrives at the result. First maybe you should find a formula for $t_n$. We are told that $t_1=1$. From the recurrence, you can conclude that $t_2=1+2$. From the recurrence, you can conclude that $t_3=(1+2)+3$, and so on. So informally you know that $t_n=1+2+3+\cdots +n$. Do you remember a formula or method for this sum? It is $n(n+1)/2$. You may be expected to prove this using any method, or perhaps you must use induction. Certainly $t_1=(1)(2)/2$. (Continued) –  André Nicolas Jan 24 '12 at 6:30

2 Answers 2

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Hint:

$$\begin{align} t_1 &= 1 \\ t_2 &= 1 + 2 \\ t_3 &= 1 + 2 + 3 \\ t_4 &= 1 + 2 + 3 + 4 \end{align}$$

Can you guess a formula for $t_n$? Once you have a formula for $t_n$, use that formula to prove that $s_n = \frac{(n+1)^3 - n - 1}{6}$.

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what does a "closed formula" mean? –  Raynos Jan 25 '12 at 5:01
    
By a "closed formula", I mean that $t_n = \frac{n(n_1)}{2}$ (which itself can be proven by induction). Once you know that $t_n = \frac{n(n+1)}{2}$, then prove that $s_n = \frac{(n+1)^3-n-1}{6}$ also by induction. –  JavaMan Jan 27 '12 at 18:11

Since : $t_n=\frac{n(n+1)}{2}$ it follows that :

$s_n=s_{n-1}+\frac{n(n+1)}{2} ; n \geq 2$ ,therefore :

$$\begin{align} s_1 &= 1 \\ s_2 &= 1 + 3 = 4 \\ s_3 &= 4+6=10 \\ s_4 &= 10+10=20 \\ s_5 &= 20+15=35 \end{align}$$

Now, note that : $s_n = t_n \cdot \frac{n+2}{3}$ and use fact that : $(n+1)^3-n-1=n(n+1)(n+2)$

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