Question

What is \[\sum\limits_{r = 0}^1 {{}^{n + r}{C_n}} {\text{ }}\] equal to ?
A. \[{}^{n + 2}{C_1}\]
B. \[{}^{n + 2}{C_n}\]
C. \[{}^{n + 3}{C_n}\]
D. \[{}^{n + 2}{C_{n - 1}}\]

Answer 1

Hint: For solving this particular question, we have to solve \[\sum\limits_{r = 0}^1 {{}^{n + r}{C_n}} {\text{ }}\], firstly we have to solve the summation then the combination. We can simplify the terms with the help of certain combination rules such as $[{}^n{C_r} = {}^n{C_{n - r}}]$ and ${}^n{C_{r - 1}} + {}^n{C_r} = {}^{n + 1}{C_r}$ . The order or the arrangement of objects in a combination does not count. It is just the selection or the inclusion of objects which is essential, and not its arrangement with respect to the other selected objects.

Formula Used:
We have used the following relation ,
$[{}^n{C_r} = {}^n{C_{n - r}}]$ ,
and ${}^n{C_{r - 1}} + {}^n{C_r} = {}^{n + 1}{C_r}$ .

Complete step by step solution:
Since $[{}^n{C_r} = {}^n{C_{n - r}}]$
and ${}^n{C_{r - 1}} + {}^n{C_r} = {}^{n + 1}{C_r}$
\[
  \begin{array}{*{20}{l}}
  {\sum\limits_{r = 0}^1 {{}^{n + r}{C_n}} {\text{ }} = \sum\limits_{r = 0}^1 {{}^{n + r}{C_r}} {\text{ }} = {\text{ }}{}^n{C_0} + {}^{n + 1}{C_1}} \\
  { = \left[ {1 + \left( {n + 1} \right)} \right] + {}^{n + 2}{C_2} + {}^{n + 3}{C_3} + ....... + {}^{n + 1}{C_1}} \\
  { = {\text{ }}\left( {{}^{n + 1}{C_1} + {}^{n + 2}{C_2}} \right) + {}^{n + 3}{C_3} + ....... + {}^{n + 1}{C_1}} \\
  { = \because n + 2 = {}^{n + 1}{C_1}{\text{ }}or{\text{ }}{}^n{C_1} = n} \\
  { = {\text{ }}\left( {{}^{n + 3}{C_2} + {}^{n + 3}{C_3}} \right) + ....... + {}^{n + 1}{C_1}} \\
  { = {\text{ }}\left( {{}^{n + 4}{C_3} + {}^{n + 4}{C_4}} \right) + ....... + {}^{n + 1}{C_1}} \\
  {...........................................................................................} \\
  {...........................................................................................} \\
  { = {\text{ }}{}^{n + 1}{C_0} + {}^{n + 1}{C_1}}
\end{array} \\
   = {}^{n + 2}{C_1} = {}^{n + 2}{C_{n + 1}}\;{\text{ }}[\because {}^n{C_r} = {}^n{C_{n - r}}]\;{\text{ }}\;{\text{ }}\; \\
 \]
Hence we can say that \[\sum\limits_{r = 0}^1 {{}^{n + r}{C_n}} {\text{ }}\]equals to ${}^n{C_{r - 1}} + {}^n{C_r} = {}^{n + 1}{C_r}$ .

Therefore, we can say that option ‘A’ is the correct option.

Additional Information:
Each of the different groups or you can say selections which can be made by some or all of the number of given things without reference to the order of the things in each group is known as combination.
The number of combinations of $n$ different things taken $r$ at a time is given by ${}^n{C_r}$ or $C(n,r)$ or $(nr)$ .
Then, ${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
Or $ = \dfrac{{{}^n{P_r}}}{{r!}}$
The order or the arrangement of objects in a combination does not count. It is just the selection or the inclusion of objects which is essential, and not its arrangement with respect to the other selected objects.
The number of ways of selecting $r$ different things out of $n$different things is $n + 1$ .

Note: We are not required to expand ${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$ , we just have to manipulate terms in the combination form only. If we have questions similar in nature as that of above can be approached in a similar manner and we can solve it easily and can find the corresponding result.