Question

The sum of the numbers $1 + 2 \cdot 2 + 3 \cdot {2^2} + 4 \cdot {2^3} + ..... + 50 \cdot {2^{49}}$ is
(1) $1 + 49 \cdot {2^{49}}$
(2) $1 + 49 \cdot {2^{50}}$
(3) $1 + 50 \cdot {2^{49}}$
(4) $1 + 50 \cdot {2^{50}}$

Answer 1

To solve this problem, we let the given sum of numbers as $S$ and named is as equation $\left( i \right)$ then we multiply the equation $\left( i \right)$ by $2$ and named it as equation $\left( {ii} \right)$ and we write it under equation $\left( i \right)$ in such a way that the right side of the equation $\left( {ii} \right)$ starts under the second term of the equation $\left( i \right)$ Then subtract both the equations. After some simplification, we get a G.P and we will apply the formula of sum of n terms of G.P. and do the calculation and find the desired result.
Formula used:
${S_n} = a\left( {\dfrac{{{r^n} - 1}}{{r - 1}}} \right)$
where, a is the first term, r is the common ratio and n is the number of terms.

Complete answer: Let,
$S = 1 + 2 \cdot 2 + 3 \cdot {2^2} + 4 \cdot {2^3} + 5 \cdot {2^4} + ..... + 50 \cdot {2^{49}}{\text{ }} - - - \left( i \right)$
Multiply above equation by $2$ we get
$2S = {\text{ }}1 \cdot 2 + 2 \cdot {2^2} + 3 \cdot {2^3} + 4 \cdot {2^4} + ...... + 49 \cdot {2^{49}} + 50 \cdot {2^{50}}{\text{ }} - - - \left( {ii} \right)$
Now, subtract $\left( {ii} \right)$ from $\left( i \right)$
$- S = 1 + 1 \cdot 2 + 1 \cdot {2^2} + 1 \cdot {2^3} + 1 \cdot {2^4} + .... + 1 \cdot {2^{49}} - 50 \cdot {2^{50}}$
$\Rightarrow - S = 1 + \left( {2 + {2^2} + {2^3} + {2^4} + .... + {2^{49}}} \right) - 50 \cdot {2^{50}}{\text{ }} - - - \left( {iii} \right)$
Now, the terms that are under bracket forms a G.P. So, we will apply the formula of sum of n terms of G.P.
i.e., ${S_n} = a\left( {\dfrac{{{r^n} - 1}}{{r - 1}}} \right)$
Here, $a = 2,{\text{ }}n = 49,{\text{ }}r = 2$
$\therefore {S_{49}} = 2\left( {\dfrac{{{2^{49}} - 1}}{{2 - 1}}} \right)$
$\Rightarrow {S_{49}} = 2\left( {{2^{49}} - 1} \right)$
Substitute the value of ${S_{49}}$ in equation $\left( {iii} \right)$
$\Rightarrow - S = 1 + 2\left( {{2^{49}} - 1} \right) - 50 \cdot {2^{50}}$
$\Rightarrow - S = 1 + ({2^{50}} - 2) - 50 \cdot {2^{50}}$
On simplification, we get
$\Rightarrow - S = - 1 - 49 \cdot {2^{50}}$
$\Rightarrow S = 1 + 49 \cdot {2^{50}}$
Hence, the sum of the numbers $1 + 2 \cdot 2 + 3 \cdot {2^2} + 4 \cdot {2^3} + ..... + 50 \cdot {2^{49}}$ is $1 + 49 \cdot {2^{50}}$
Hence, option $\left( 2 \right)$ is correct.

Note:
This type of series is known as Arithmetic geometric series (A.G.S). Its ${{\text{n}}^{{\text{th}}}}$ can be written as ${T_n} = n\left( {{2^n} - 1} \right)$ with ${\text{n}}$ terms as A.P. and $\left( {{2^n} - 1} \right)$ terms as G.P. In this question series represents A.G.S with A.P as $1,2,3,...50$ and G.P. as $2,{2^2},{2^3},...,{2^{49}}$ . Also, while solving these types of questions the idea of the correct answer can be gained by looking at the options. Like in this question if we observe there is a pattern in the series. i.e.,
$S\left( 1 \right) = 1$
$S\left( 2 \right) = 1 + 2 \cdot 2 = 1 + {2^2}$
$S\left( 3 \right) = 1 + 2 \cdot 2 + {3.2^2} = 1 + 4 + 12 = 17 = 1 + 2 \cdot {2^3}$
and so on
So, $S\left( n \right) = 1 + \left( {n - 1} \right){2^n}$
$\therefore S\left( {50} \right) = 1 + \left( {50 - 1} \right){2^{50}} = 1 + 49 \cdot {2^{50}}$ which is the correct answer.
Hence, option $\left( 2 \right)$ will be correct.