Sequence Within A Sequence 3

We claim that $T_n = \frac{1}{2}n(2315-3n)$ and prove it by induction.

1. For $n=1$ , $T_1 = \frac{1}{2}(1)(2315-3(1)) = 1156$ as given. The claim is true for $n=1$ .
2. Assume that the claim is true for $n$ , then

$\begin{aligned} \quad \quad T_{\color{#D61F06}{n+1}} & = T_n + \color{#3D99F6}{d_n} \quad \quad \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{We note that }d_n = d_1 - 3(n-1) = 1156-3n} \\ & = \frac{n(2315-3n)}{2} + \color{#3D99F6}{1156-3n} \\ & = \frac{2315n-3n^2 + 2312-6n}{2} \\ & = \frac{2312+2309n-3n^2}{2} \\ & = \frac{(\color{#D61F06}{n+1})(2315-3(\color{#D61F06}{n+1}))}{2} \end{aligned}$

$\quad$ The claim is also true for $n+1$ and therefore it is true for all $n \ge 1$ .

Now, we have:

$\begin{aligned} S_n & = 0 \\ \sum_{k=1}^n T_k & = 0 \\ \sum_{k=1}^n \frac{2315k-3k^2}{2} & = 0 \\ \frac{2315n(n+1)}{4} - \frac{3n(n+1)(2n+1)}{12} & = 0 \\ \Rightarrow 2315n(n+1) & = n(n+1)(2n+1) \\ 2n + 1 & = 2315 \\ \Rightarrow n & = \boxed{1157} \end{aligned}$

From Sequence Within A Sequence 2 , we know that $S_n=nT_1+\frac {n (n-1)[3d_1+(n-2)D]}{6}$ Plugging in $T_1=1156$ , $d_1=1153$ , $D=-3$ , $S_n=0$ , we have $0=1156n+\frac {n (n-1)[3×1153-3 (n-2)]}{6}$ $\frac {(n-1)(n-1155)}{2}-1156=0$ $\frac {n^2-1156n+1155}{2}-1156=0$ $n^2-1156n-1157=(n+1 )(n-1157)=0$ Since $n>0$ , thus $n=1157$ .