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

Hint: 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 \[1\].