Question

The value of is equal to \[\sum\limits_{n=0}^{1947}{\dfrac{1}{\mathop{2}^{n}+\sqrt{\mathop{2}^{1947}}}}\]
A) \[\dfrac{487}{\sqrt{\mathop{2}^{1945}}}\]
B) \[\dfrac{1946}{\sqrt{\mathop{2}^{1947}}}\]
C) \[\dfrac{1947}{\sqrt{\mathop{2}^{1947}}}\]
D) \[\dfrac{1948}{\sqrt{\mathop{2}^{1947}}}\]

Answer 1

Hint: In this type of question we always write series in expanded form first and then write the same series in reverse form. After that we can add both forms of series and get a desired expression.

Complete step by step solution:
Consider\[\sum\limits_{n=0}^{1947}{\dfrac{1}{\mathop{2}^{n}+\sqrt{\mathop{2}^{1947}}}}\]
We can write it as
\[x=\sum\limits_{n=0}^{1947}{\dfrac{1}{\mathop{2}^{n}+{{2}^{\dfrac{1947}{2}}}}}\]………………………………………………….(i)
On expanding
 \[x=\sum\limits_{n=0}^{1947}{\dfrac{1}{\mathop{2}^{n}+\mathop{2}^{\dfrac{1947}{2}}}=\dfrac{1}{\mathop{2}^{0}+\mathop{2}^{\dfrac{1947}{2}}}+\dfrac{1}{\mathop{2}^{1}+\mathop{2}^{\dfrac{1947}{2}}}}+.................+\dfrac{1}{\mathop{2}^{1947}+\mathop{2}^{\dfrac{1947}{2}}}\] ……………………(ii)
Reverse the equation (ii)
\[x=\dfrac{1}{\mathop{2}^{1947}+\mathop{2}^{\dfrac{1947}{2}}}+\dfrac{1}{\mathop{2}^{1946}+\mathop{2}^{\dfrac{1947}{2}}}+.................+\dfrac{1}{\mathop{2}^{1}+\mathop{2}^{\dfrac{1947}{2}}}+\dfrac{1}{\mathop{2}^{0}+\mathop{2}^{\dfrac{1947}{2}}}\]
Based on above pattern we can write this
\[\Rightarrow x=\sum\limits_{n=0}^{1947}{\dfrac{1}{\mathop{2}^{1947-n}+\mathop{2}^{\dfrac{1947}{2}}}}\]…………………………………………………….(iii)
On adding equation (ii) and (ii)
\[\Rightarrow x+x=\sum\limits_{n=0}^{1947}{\dfrac{1}{\mathop{2}^{n}+\mathop{2}^{\dfrac{1947}{2}}}}+\sum\limits_{n=0}^{1947}{\dfrac{1}{\mathop{2}^{1947-n}+\mathop{2}^{\dfrac{1947}{2}}}}\]
\[\Rightarrow 2x=\sum\limits_{n=0}^{1947}{\left( \dfrac{1}{\mathop{2}^{n}+\mathop{2}^{\dfrac{1947}{2}}}+\dfrac{1}{\mathop{2}^{1947-n}+\mathop{2}^{\dfrac{1947}{2}}} \right)}\]
From the above expression we can take \[\mathop{2}^{n}\]as common from first term and \[\mathop{2}^{\dfrac{1947}{2}}\]as common from second term in denominator.
\[\Rightarrow 2x=\sum\limits_{n=0}^{1947}{\left( \dfrac{1}{{{2}^{n}}\left( 1+{{2}^{\dfrac{1947}{2}-n}} \right)}+\dfrac{1}{{{2}^{\dfrac{1947}{2}}}\left( {{2}^{1947-n-\dfrac{1947}{2}}}+1 \right)} \right)}\]
\[\Rightarrow 2x=\sum\limits_{n=0}^{1947}{\left( \dfrac{1}{{{2}^{n}}\left( 1+{{2}^{\dfrac{1947}{2}-n}} \right)}+\dfrac{1}{{{2}^{\dfrac{1947}{2}}}\left( {{2}^{\dfrac{1947}{2}-n}}+1 \right)} \right)}\]
\[\Rightarrow 2x=\sum\limits_{n=0}^{1947}{\left( \dfrac{1}{\left( 1+{{2}^{\dfrac{1947}{2}-n}} \right)}\left( \dfrac{1}{{{2}^{n}}}+\dfrac{1}{{{2}^{\dfrac{1947}{2}}}} \right) \right)}\]
\[\Rightarrow 2x=\sum\limits_{n=0}^{1947}{\left( \dfrac{1}{\left( 1+{{2}^{\dfrac{1947}{2}-n}} \right)}\left( \dfrac{{{2}^{\dfrac{1947}{2}-n}}+1}{{{2}^{\dfrac{1947}{2}}}} \right) \right)}\]
\[\Rightarrow 2x=\sum\limits_{n=0}^{1947}{\left( \dfrac{1}{{{2}^{\dfrac{1947}{2}}}} \right)}\]
\[\Rightarrow 2x=\dfrac{1948}{{{2}^{\dfrac{1947}{2}}}}\]
\[\Rightarrow x=\dfrac{974}{{{2}^{\dfrac{1947}{2}}}}\]
\[\Rightarrow x=\dfrac{487}{{{2}^{\dfrac{1945}{2}}}}\]
\[\Rightarrow x=\dfrac{487}{\sqrt{{{2}^{1945}}}}\]
Hence option (a) is correct.

Note: These types of problems are called sequence and series problems. You have to try solving these sums by making small amounts of changes and try to use progression formulas.