Sequence Within A Sequence 4

First few values of $T_n$ show us that $T_n = 4+2^{n-1}$ , but is it true for all value of $n \ge 1$ ? Let us prove it by induction.

We claim that $T_n = 4+2^{n-1}$ .

1. For $n=1$ , $T_1 = 4 + 2^{1-1} = 5$ , which is as given. So the claim is true for $n=1$ .
2. Now assume that the claim is true for $n$ , then we have:

$\begin{aligned} T_{\color{#D61F06}{n+1}} & = T_n + \color{#3D99F6}{d_n \quad \quad \quad \quad \quad \quad \quad \small \text{We note that } d_n = 2^{n-1}} \\ & = 4+2^{n-1} + \color{#3D99F6}{2^{n-1}} \\ & = 4 + 2^n \\ & = 4 + 2^{\color{#D61F06}{n+1}-1} \end{aligned}$

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

Now, we have $T_{16} = 4 + 2^{16-1} = \boxed{32772}$

$d_k = 2\times d_{k-1} = 2\times 2\times d_{k-2} = 2\times 2 \times \ldots ( (k-1) times ) \times d_1= 2^{k-1}$

$T_n = T_{n-1} + d_{n-1}$

$T_n = T_{n-2} + d_{n-2} + 2^{n-2}$

$T_n = T_1 + d_1 + 2^1 + 2^2 + 2^3 + \ldots + 2^{n-2}$

$T_n = 5 + \frac{2^{n-1} - 1}{2-1}$

$T_n = 5 + 2^{n-1} - 1$

$T_n = 4 + 2^{n-1}$

$\displaystyle \Rightarrow T_{16} = 4 + 2^{15} = \boxed{32772}$

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 a G.P. (Geometric Progression).

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

Now, substitute $n=16$ , $T_1=5$ , $d_1=1$ , $R=2$ , we could get $T_{16}=32772$ .