Question

The sum of the series $2{}^{20}{{C}_{0}}+5{}^{20}{{C}_{1}}+8{}^{20}{{C}_{2}}+11{}^{20}{{C}_{3}}+\cdots +62{}^{20}{{C}_{20}}$ is equal to:
(a) ${{2}^{24}}$
(b) ${{2}^{25}}$
(c) ${{2}^{26}}$
(d) ${{2}^{23}}$

Answer 1

Hint: We will first rearrange the series $2{}^{20}{{C}_{0}}+5{}^{20}{{C}_{1}}+8{}^{20}{{C}_{2}}+11{}^{20}{{C}_{3}}+\cdots +62{}^{20}{{C}_{20}}$ into a similar expression and then we will try to find its sum with the original series. Then after using the formula $\sum\limits_{0\le k\le n}{{}^{n}{{C}_{k}}}={{2}^{n}}$ of the sum of the nth row, we will try to find the sum of the series.

Complete step-by-step solution:
We know from the properties of combinations that ${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$
Then, for the terms given in the expression mentioned in the question,
 $\begin{align}
  &\Rightarrow {}^{20}{{C}_{0}}={}^{20}{{C}_{20-0}}={}^{20}{{C}_{20}} \\
 &\Rightarrow {}^{20}{{C}_{1}}={}^{20}{{C}_{20-1}}={}^{20}{{C}_{19}} \\
 &\Rightarrow {}^{20}{{C}_{2}}={}^{20}{{C}_{20-2}}={}^{20}{{C}_{18}} \\
 &\Rightarrow {}^{20}{{C}_{3}}={}^{20}{{C}_{20-3}}={}^{20}{{C}_{17}} \\
\end{align}$
And so on...
We should also recall that $\sum\limits_{0\le k\le n}{{}^{n}{{C}_{k}}}={{2}^{n}}$
\[\Rightarrow {}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}+{}^{n}{{C}_{3}}+{}^{n}{{C}_{4}}+\cdots +{}^{n}{{C}_{n}}={{2}^{n}}\text{ }\cdots \left( i \right)\]
Let us write the sum as $S=2{}^{20}{{C}_{0}}+5{}^{20}{{C}_{1}}+8{}^{20}{{C}_{2}}+11{}^{20}{{C}_{3}}+\cdots +62{}^{20}{{C}_{20}}$
Then it can also be written as $S=2{}^{20}{{C}_{20}}+5{}^{20}{{C}_{19}}+8{}^{20}{{C}_{18}}+11{}^{20}{{C}_{17}}+\cdots +62{}^{20}{{C}_{0}}$
This can also be arranged as $S=62{}^{20}{{C}_{20}}+59{}^{20}{{C}_{19}}+\cdots +2{}^{20}{{C}_{0}}$
 Two rearrangements of S can be found as
$\begin{align}
  & S=2{}^{20}{{C}_{0}}+5{}^{20}{{C}_{1}}+8{}^{20}{{C}_{2}}+11{}^{20}{{C}_{3}}+\cdots +62{}^{20}{{C}_{20}} \\
 & S=62{}^{20}{{C}_{20}}+59{}^{20}{{C}_{19}}+\cdots +2{}^{20}{{C}_{0}} \\
\end{align}$
Adding them, we get
$\begin{align}
  &\Rightarrow S+S=\left( 2+62 \right){}^{20}{{C}_{20}}+\left( 5+59 \right){}^{20}{{C}_{19}}+\cdots +\left( 2+62 \right){}^{20}{{C}_{0}} \\
 &\Rightarrow 2S=64{}^{20}{{C}_{20}}+64{}^{20}{{C}_{19}}+\cdots +64{}^{20}{{C}_{0}} \\
 & =64\left( {}^{20}{{C}_{20}}+{}^{20}{{C}_{19}}+\cdots +{}^{20}{{C}_{0}} \right)
\end{align}$
Putting the value of equation (i) in above equation, we get
$\begin{align}
  &\Rightarrow 2S=64\left( {}^{20}{{C}_{20}}+{}^{20}{{C}_{19}}+\cdots+{}^{20}{{C}_{0}} \right) \\
 &\Rightarrow 2S=64\times {{2}^{20}} \\
 &\Rightarrow S=32\times {{2}^{20}} \\
 & ={{2}^{5}}\times {{2}^{20}}={{2}^{25}}
\end{align}$
Hence, option (c) is correct.

Note: Let us recall the definition of factorial notation and combinations.
The factorial notation is denoted by $n!$ or $\left| \!{\underline {\,
  n \,}} \right. $ and is defined as the product of first n natural numbers, that is, $n!=1\times 2\times 3\times \cdots \times n$
The number of combinations of n different objects taken r at a time, is denoted by ${}^{n}{{C}_{r}}$ and is defined by ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Another way to solve this question is,
Given $S=2{}^{20}{{C}_{0}}+5{}^{20}{{C}_{1}}+8{}^{20}{{C}_{2}}+11{}^{20}{{C}_{3}}+\cdots +62{}^{20}{{C}_{20}}$
Clearly, 2,5,8,... follows the pattern of 3r+2 where r=0, 1, 2, ...
Thus, the given series can be written as
$S=2{}^{20}{{C}_{0}}+5{}^{20}{{C}_{1}}+8{}^{20}{{C}_{2}}+11{}^{20}{{C}_{3}}+\cdots +62{}^{20}{{C}_{20}}$
Putting $3r+2$ in the above expression, we get
$S=\sum\limits_{r=0}^{20}{\left( 3r+2 \right)}{}^{20}{{C}_{r}}$
Solving it further, we get
$\begin{align}
  & S=\sum\limits_{r=0}^{20}{3r\left( {}^{20}{{C}_{r}} \right)}+\sum\limits_{r=0}^{20}{2\left( {}^{20}{{C}_{r}} \right)} \\
 & =3\sum\limits_{r=0}^{20}{r\left( {}^{20}{{C}_{r}} \right)}+2\sum\limits_{r=0}^{20}{\left( {}^{20}{{C}_{r}} \right)}
\end{align}$
We know, ${}^{n}{{C}_{r}}=\dfrac{n}{r}{}^{n-1}{{C}_{r-1}}$
Putting this value in above expression, we get
$S=3\sum\limits_{r=0}^{20}{r\left( \dfrac{20}{r} \right)\left( {}^{19}{{C}_{r-1}} \right)}+2\sum\limits_{r=0}^{20}{\left( {}^{20}{{C}_{r}} \right)}$
Solving it further, we get
$\begin{align}
  & S=3\times 20\sum\limits_{r=0}^{20}{\left( {}^{19}{{C}_{r-1}} \right)}+2\sum\limits_{r=0}^{20}{\left( {}^{20}{{C}_{r}} \right)} \\
 & =60\times {{2}^{19}}+2\times {{2}^{20}} \\
 & =\left( 60+4 \right)\times {{2}^{19}}=64\times {{2}^{19}}={{2}^{6}}\times {{2}^{19}}={{2}^{25}}
\end{align}$
Hence, the sum of the series is ${{2}^{25}}$.