recursive relation with sequences

Question

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.

Answer 1

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 $s_n = \frac{(n+1)^3 - n - 1}{6}$.

Answer 2

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 : $(n+1)^3-n-1=n(n+1)(n+2)$