Question

The sum of the series \[\sum\limits_{r=0}^{10}{{}^{20}{{C}_{r}}}\] is: -
(a) \[{{2}^{20}}\]
(b) \[{{2}^{19}}\]
(c) \[{{2}^{19}}+\dfrac{1}{2}{}^{20}{{C}_{10}}\]
(d) \[{{2}^{19}}-\dfrac{1}{2}{}^{20}{{C}_{10}}\]

Answer 1

Hint:
 Assume the sum of the given expression as ‘E’. Multiply the given expression with 2 and then divide it by 2 to balance the expression. Use the formula: - \[{}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}\] after writing the expression in the expanded form by removing the summation sign. Use the above formula for only half of the terms. Now, simplify the terms by using the formula: - \[{}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}+.......+{}^{n}{{C}_{n}}={{2}^{n}}\] to get the correct answer.

Complete step by step answer:
Here, we have been asked to find the sum of the series, \[\sum\limits_{r=0}^{10}{{}^{20}{{C}_{r}}}\].
Now, let us assume the required sum of the expression as ‘E’. So, we have,
\[\Rightarrow E=\sum\limits_{r=0}^{10}{{}^{20}{{C}_{r}}}\]
Multiplying the given expression with 2 and then dividing it by 2 to balance it, we get,
\[\Rightarrow E=\dfrac{1}{2}\times \left( 2\sum\limits_{r=0}^{10}{{}^{20}{{C}_{r}}} \right)\]
Expanding the sum by removing the summation sign we get,
\[\Rightarrow E=\dfrac{1}{2}\times \left\{ 2\left( {}^{20}{{C}_{0}}+{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}+.......+{}^{20}{{C}_{10}} \right) \right\}\]
The above expression can be written as: -
\[\Rightarrow E=\dfrac{1}{2}\times \left\{ \left( {}^{20}{{C}_{0}}+{}^{20}{{C}_{1}}+.......+{}^{20}{{C}_{10}} \right)+\left( {}^{20}{{C}_{0}}+{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}+.......+{}^{20}{{C}_{10}} \right) \right\}\]
We can clearly see that in the above sum, two similar expressions are present inside the bracket. So, leaving one of these expressions as it is and applying the identity, \[{}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}\] for the other one, we get,
\[\Rightarrow E=\dfrac{1}{2}\times \left\{ \left( {}^{20}{{C}_{0}}+{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}+.......+{}^{20}{{C}_{10}} \right)+\left( {}^{20}{{C}_{20}}+{}^{20}{{C}_{19}}+{}^{20}{{C}_{18}}+.......+{}^{20}{{C}_{10}} \right) \right\}\]
Rearranging the terms inside the bracket, we can write the above series as: -
\[\Rightarrow E=\dfrac{1}{2}\times \left\{ \left( {}^{20}{{C}_{0}}+{}^{20}{{C}_{1}}+{}^{20}{{C}_{2}}+.......+{}^{20}{{C}_{20}} \right)+{}^{20}{{C}_{10}} \right\}\]
Now, using the formula: - \[{}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}+.......+{}^{n}{{C}_{n}}={{2}^{n}}\], we get,
\[\begin{align}
  & \Rightarrow E=\dfrac{1}{2}\left\{ {{2}^{20}}+{}^{20}{{C}_{10}} \right\} \\
 & \Rightarrow E=\dfrac{{{2}^{20}}}{2}+\dfrac{1}{2}\left( {}^{20}{{C}_{10}} \right) \\
\end{align}\]
Using the formula: - \[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\], we get,
\[\begin{align}
  & \Rightarrow E={{2}^{20-1}}+\dfrac{1}{2}\left( {}^{20}{{C}_{10}} \right) \\
 & \Rightarrow E={{2}^{19}}+\dfrac{1}{2}\left( {}^{20}{{C}_{10}} \right) \\
\end{align}\]
Therefore, the sum of the given series is \[{{2}^{19}}+\dfrac{1}{2}\left( {}^{20}{{C}_{10}} \right)\].
Hence, option (c) is the correct answer.

Note:
One may note that we do not have to use the formula \[{}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}\] for all the terms inside the bracket otherwise this formula will be of no use. So, change only half of the terms and simplify them. Remember the formula \[\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}={{2}^{n}}\] to solve the question. Note that there are no shortcut methods to solve this question so you must remember the step – by – step procedure to solve it. The question may be asked with different limits of r but the approach will be the same. The formulas used in the solution are one of the most important formulas in combinations, so you must memorize them.