Question

If $S_1=\left\{2\right\},S_2=\{3,6\},S_3=\{4,8,16\},S_4=\{5,10,20,40\},...$ then the sum of numbers in the set $S_{15}$ is

• A. $5\left(2^{15}\right)$
• B. $16(2^{15}-1)$
• C. $16(2^{16}-1)$
• D. $15(2^{15}-1)$

Answer 1

Answer: (B)

$16(2^{15}-1)$

$S_1=\left\{2\right\}$

$S_2=\{3,6\}$

$S_3=\{4,8,16\}$

$S_4=\{5,10,20,40\}$

looking at the trend we can write a general set $S_n$ as :

$\Rightarrow S_n=\{2^0(n+1),2(n+1),2^2(n+1),....{.2}^{n-1}(n+1)\}$

the elements in $S_{15}$ set can be obtained by a substituting $n=15$ in the above equation.

$\Rightarrow S_{15}=\{\left(16\right),2\left(16\right),2^2\left(16\right),....{.2}^{14}\left(16\right)\}$

Sum of the elements :

$=2^0\left(16\right),2^1\left(16\right),2^2\left(16\right),....{.2}^{14}\left(16\right)$

$=16\{2^0+2^1+2^2+.....+2^{14}\}$

$=16\{1+2+2^2+.....{.2}^{14}\}$

we know that the sum of an GP containing n terms is given by :

$\Rightarrow S_n=\frac{a(r^n-1)}{r-1};\left|r\right|>1$ , where a is the first term and 'r' is the common ratio.

Using this,

$=16\{1+2+2^2+......{.2}^{14}\}$

$=16\frac{(2^{15}-1)}{2-1}$

$=16(2^{15}-1)$

Hence, the answer is $16(2^{15}-1).$