Question 458 of 519MediumMCQ
The sum
\[\sum\limits_{k = 1}^{20} {k{1 \over {{2^k}}}} \] is equal to
Solutions · 1
Official solution
Let S = \[\sum\limits_{k = 1}^{20} {k{1 \over {{2^k}}}} \]
\[ \Rightarrow \] S = \[{1 \over 2} + {2 \over {{2^2}}} + {3 \over {{2^3}}} + {4 \over {{2^4}}} + ....... + {{20} \over {{2^{20}}}}\] .......(1)
This is an Arithmetic Geometric Sequence. Here numerator is in A.P and denominator is in G.P.
To solve Arithmetic Geometric Sequence, we multiply the series S with the common ratio of G.P. Here common ratio of G.P is \[{1 \over 2}\].
By multiplying (1) with \[{1 \over 2}\] we get,
\[{S \over 2} = {1 \over {{2^2}}} + {2 \over {{2^3}}} + {3 \over {{2^4}}} + {4 \over {{2^5}}} + ....... + {{19} \over {{2^{20}}}} + {{20} \over {{2^{21}}}}\] ....(2)
Subtract (2) from (1),
S - \[{S \over 2}\] = \[{1 \over 2} + {1 \over {{2^2}}} + {1 \over {{2^3}}} + {1 \over {{2^4}}} + ....... + {1 \over {{2^{20}}}} - {{20} \over {{2^{21}}}}\]
\[ \Rightarrow \] \[{S \over 2}\] = \[{{{1 \over 2}\left( {1 - {1 \over {{2^{20}}}}} \right)} \over {1 - {1 \over 2}}}\] - \[{{20} \over {{2^{21}}}}\]
\[ \Rightarrow \] S = \[2\left( {1 - {1 \over {{2^{20}}}}} \right) - {{20} \over {{2^{20}}}}\]
\[ \Rightarrow \] S = \[2 - {2 \over {{2^{20}}}} - {{20} \over {{2^{20}}}}\]
\[ \Rightarrow \] S = \[2 - {{22} \over {{2^{20}}}}\]
\[ \Rightarrow \] S = \[2 - {{11} \over {{2^{19}}}}\]
Answer the question to unlock