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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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    Suppose that for some $k$, we know $t_k=k(k+1)/2$. Then $t_{k+1}=k(k+1)/2 +(k+1)$, which simplifies to $(k+1)(k+2)/2$, the right thing. There are other ways o do the work. For example, let $w_n=n(n+1)/2$. Show that the $w_n$ satisfy the same recurrence as the $t_n$, and start the same way, and therefore $w_n=t_n$ for all $n$.2012-01-24

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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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    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.2012-01-27
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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 : $(n+1)^3-n-1=n(n+1)(n+2)$