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

Hint: 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.