Question

The positive integer $n$ for which $2 \times {2^2} + 3 \times {2^3} + 4 \times {2^4} + ...... + n \times {2^n} = {2^{n + 10}}$ is _______
A.$510$
B.$511$
C.$512$
D.$513$

Answer 1

We have given a series for positive integers $n$. We have also given their sum. We have to find the value of $n$. Firstly, we put the series equal to $Sn$, then name it as equation (1). After that, we multiply this equation by $2$ and name it as equation (2). We will subtract equation (1) by equation (2) and solve for $n$.

Complete step-by-step answer:
We have given that
$2 \times {2^2} + 3 \times {2^3} + 4 \times {2^4} + ...... + n \times {2^n}$
$\Rightarrow 2 \times {2^2} + 3 \times {2^3} + 4 \times {2^4} + ...... + n \times {2^n} = {2^{n + 10}}$
We have to find the value of $n$
Let $Sn = 2 \times {2^2} + 3 \times {2^3} + 4 \times {2^4} + ...... + \left( {n - 1} \right) \times {2^{n - 1}} + n \times {2^n}$ ……..(1)
Multiplying equation (1) by 2
$\Rightarrow 2Sn = 2 \times {2^3} + 3 \times {2^4} + 4 \times {2^5} + ...... + \left( {n - 1} \right) \times {2^n} + n \times {2^{n + 1}}$ …….(2)
Subtracting equation (2) from equation (1), we get
$- Sn = 2 \times {2^2} + {2^3} + {2^4} + ...... + {2^n} - n \times {2^{n + 1}}$
Now, $Sn$ is equal to ${2^{n + 10}}$
So, $- {2^{n + 10}} = 2 \times {2^2} + {2^3} + {2^4} + ...... + {2^n} - n \times {2^{n + 1}}$
$a$
Here, ${2^3} + {2^4} + {2^5} + ...... + {2^n}$ forms the geometric progression whose first term is ${2^3}$ and total number of terms are $n - 2$.
Now, sum of $n$ terms of geometric progression is given as $a\left( {\dfrac{{1 - {r^n}}}{{1 - r}}} \right)$ where $a$ is first term and $r$ is common ratio.
So, ${2^3} + {2^4} + {2^5} + ...... + {2^n} = {2^3}\left( {\dfrac{{1 - {2^{n - 2}}}}{{1 - 2}}} \right)$
Therefore, $- {2^{n + 10}} = 8 + {2^3}\left( {\dfrac{{1 - {2^{n - 1}}}}{{1 - 2}}} \right) - n \times {2^{n + 1}}$
$- {2^{n + 10}} = 8 + 8\left( {\dfrac{{1 - {2^{n - 2}}}}{{ - 1}}} \right) - n \times {2^{n + 1}}$
$\Rightarrow$ $- {2^{n + 10}} = 8 - 8\left( {1 - {2^{n - 2}}} \right) - n \times {2^{n + 1}}$
Multiplying $8$ by bracket, we get
$\Rightarrow$ $- {2^{n + 10}} = 8 - 8 + 8 \times {2^{n - 2}} - n \times {2^{n + 1}}$
$\Rightarrow$ $- {2^{n + 10}} = {2^3} \times {2^{n - 2}} - n \times {2^{n + 2}}$
$\Rightarrow$ $- {2^{n + 10}} = {2^{n + 1}} - n \times {2^{n + 2}}$
Taking ${2^{n + 1}}$ common from right hand side
$\Rightarrow - {2^{n + 10}} = {2^{n + 1}}\left( {1 - n} \right)$
$\Rightarrow$ ${2^{n + 10}} = \left( {n - 1} \right) \times {2^{n + 1}}$
$\Rightarrow$ ${2^{n + 10}}$ can be written as ${2^9}{.2^{n + 1}}$
So, ${2^9}{.2^{n + 1}} = \left( {n - 1} \right) \times {2^{n + 1}}$
Cancel out the same term from the both sides
$\Rightarrow$ $n - 1 = {2^9}$
Take $- 1$ to the R.H.S.
$\Rightarrow$ $n = {2^9} + 1$
$\Rightarrow$ $n = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 + 1$
Solve this by multiplication
$\Rightarrow$ $n = 513$
So, option (D) is correct.

Note: A sequence of numbers is called geometric progression if each term after the first is written by multiplying the previous one by a fixed non-zero number. This fixed non-zero number is called the common ratio. The common ratio may be positive, negative. If $a$ is the first term of the geometric progression and $r$ is the common ratio then the geometric sequence is given as $a,ar,a{r^2},a{r^3},a{r^4}......$
Here $r$ should not be equal to $1$. The ${n^{th}}$ term of the geometric mean is given as ${a_n} = a{r^{n - 1}}$ for every $n$ greater than equal to $1$.