Question

The ${10}^{th}$ term of the series $1+5+13+29+...$ is

• A. 2045
• B. 2035
• C. 2025
• D. 2015

Answer 1

Answer: (A)

2045

Given series

$t_1=1$

$t_2=5$

$t_3=13$

$t_4=29$

$t_2-t_1=5-1=4$

$=2^2$

$t_3-t_2=13-5=8$

$=2^3$

$t_4-t_3=29-13=16$

$=2^4$

We can write this letter as $t_n-t_{n-1}=2^n$

$\therefore (t_n-t_{n-1})+(t_{n-1}-t_{n-2})\dots \dots .+(t_2-t_1)$

$=2^n+2^{n-1}\dots \dots +2^2$

$\therefore t_n-t_1=4+...{.2}^n$

$t_n-1=4(1+2+.....{.2}^{n-2})$

$t_n=4\left(\frac{2^{n-1}-1}{2-1}\right)+1$

$t_n=2^{n+1}-4+1=2^{n+1}-3$

$\therefore t_{10}=2^{11}-3=2048-3$.