Sequence Within A Sequence

We have $T_2=T_1+d_1 \\ T_3=T_2+d_2 \\ \vdots \\ T_n=T_{n-1}+d_{n-1}$ Add all these equations together, cancelling out all the like terms we get $T_n=T_1+d_1+d_2+d_3+\ldots+d_{n-1}$ From the question, we know that $d_1, d_2, \ldots, d_{n-1}$ is an A.P. (Arithmetic Progression).

Suppose $d_k-d_{k-1}=D,~~k=2, 3, \ldots, n-1$ , using the sum of A.P. formula, we now have $T_n=T_1+\frac{n-1}{2}[2d_1+(n-2)D]$

Now, substitute $n=1153$ , $T_1=5$ , $d_1=3$ , $D=1$ , we could get $T_{1153}=666437$ .

We claim that $T_n = T_1 + (n-1)d_1 + \dfrac{(n-1)(n-2)}{2} = 5 + \dfrac{(n-1)(n+4)}{2}$ and then prove it by induction.

1. For $n = 1$ , $T_1 = 5 + \dfrac{(1-1)(1+4)}{2} = 5$ as given. The claim is true for $n=1$ .
2. Assume that the claim is true for $n$ , then we have:

$\begin{aligned} \quad \quad T_{\color{#D61F06}{n+1}} & = T_n + \color{#3D99F6}{d_n} \quad \quad \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{Note that }d_n = d_1 + n-1 = n + 2} \\ & = 5 + \dfrac{(n-1)(n+4)}{2} + \color{#3D99F6}{n + 2} \\ & = 5 + \dfrac{n^2+3n-4+2n+4}{2} \\ & = 5 + \dfrac{n^2+5n}{2} \\ & = 5 + \dfrac{n(n+5)}{2} \\ & = 5 + \dfrac{(\color{#D61F06}{n+1}-1)(\color{#D61F06}{n+1}+4)}{2} \end{aligned}$

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

Therefore, $T_{1153} = 5 + \dfrac{(1153-1)(1153+4)}{2} = 5 + 576 \times 1157 = \boxed{666437}$

The series 5, 8, 12, 17, 23, . . . expressed as $\frac{n^2 + 3n + 6}{2}$ , gives 666437 at n = 1153.