Question

Calculate the value of $\sum\limits_{r = 0}^m {{}^{n + r}{C_r}} $.
A. ${}^{n + m + 1}{C_{n + 1}}$
B. ${}^{n + m + 2}{C_n}$
C. ${}^{n + m + 3}{C_{n - 1}}$
D. None of these

Answer 1

Hint: First we will expand $\sum\limits_{r = 0}^m {{}^{n + r}{C_r}} $ by putting $r = 0,1,2, \cdots ,m$. Then calculate the value of ${}^n{C_0}$ and ${}^{n + 1}{C_1}$ by using the combination formula. By using the combination property we will add consecutive terms to calculate the value of summation.

Formula Used:
${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$
${}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}$
${}^n{C_r} = {}^n{C_{n - r}}$

Complete step by step solution:
Given summation is
$\sum\limits_{r = 0}^m {{}^{n + r}{C_r}} $
Now expand the summation by putting $r = 0,1,2, \cdots ,m$.
$ = {}^n{C_0} + {}^{n + 1}{C_1} + {}^{n + 2}{C_2} + {}^{n + 3}{C_3} + {}^{n + 4}{C_4} + \cdots + {}^{n + m}{C_m}$
Now applying the formula ${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$ in the first two terms
$ = \dfrac{{n!}}{{n!0!}} + \dfrac{{\left( {n + 1} \right)!}}{{\left( {n + 1 - 1} \right)!1!}} + {}^{n + 2}{C_2} + {}^{n + 3}{C_3} + {}^{n + 4}{C_4} + \cdots + {}^{n + m}{C_m}$
$ = \dfrac{{n!}}{{n!0!}} + \dfrac{{\left( {n + 1} \right)!}}{{n!1!}} + {}^{n + 2}{C_2} + {}^{n + 3}{C_3} + {}^{n + 4}{C_4} + \cdots + {}^{n + m}{C_m}$
Putting $0! = 1$ ; $1! = 1$ and apply the formula $r! = r\left( {r - 1} \right)!$
$ = 1 + \dfrac{{\left( {n + 1} \right) \cdot n!}}{{n!}} + {}^{n + 2}{C_2} + {}^{n + 3}{C_3} + {}^{n + 4}{C_4} + \cdots + {}^{n + m}{C_m}$
$ = 1 + \left( {n + 1} \right) + {}^{n + 2}{C_2} + {}^{n + 3}{C_3} + {}^{n + 4}{C_4} + \cdots + {}^{n + m}{C_m}$
$ = \left( {n + 2} \right) + {}^{n + 2}{C_2} + {}^{n + 3}{C_3} + {}^{n + 4}{C_4} + \cdots + {}^{n + m}{C_m}$
Apply the reverse formula of ${}^{n + 2}{C_1} = \dfrac{{\left( {n + 2} \right)!}}{{\left( {n + 2 - 1} \right)!1!}} = n + 2$
$ = {}^{n + 2}{C_1} + {}^{n + 2}{C_2} + {}^{n + 3}{C_3} + {}^{n + 4}{C_4} + \cdots + {}^{n + m}{C_m}$
$ = \left( {{}^{n + 2}{C_1} + {}^{n + 2}{C_2}} \right) + {}^{n + 3}{C_3} + {}^{n + 4}{C_4} + \cdots + {}^{n + m}{C_m}$
Apply the formula ${}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}$ in third term and fourth term
$ = {}^{n + 3}{C_2} + {}^{n + 3}{C_3} + {}^{n + 4}{C_4} + \cdots + {}^{n + m}{C_m}$
$ = \left( {{}^{n + 3}{C_2} + {}^{n + 3}{C_3}} \right) + {}^{n + 4}{C_4} + \cdots + {}^{n + m}{C_m}$
Again, apply the formula ${}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}$
$ = {}^{n + 4}{C_3} + {}^{n + 4}{C_4} + \cdots + {}^{n + m}{C_m}$
$ = \left( {{}^{n + 4}{C_3} + {}^{n + 4}{C_4}} \right) + \cdots + {}^{n + m}{C_m}$
So on
$ = {}^{n + m}{C_{m - 1}} + {}^{n + m}{C_m}$
Apply the formula ${}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}$
$ = {}^{n + m + 1}{C_m}$
Now we will apply ${}^n{C_r} = {}^n{C_{n - r}}$
$ = {}^{n + m + 1}{C_{n + m + 1 - m}}$
$ = {}^{n + m + 1}{C_{n + 1}}$

Option ‘A’ is correct

Note: To solve the question that is related to the combination, you must be aware of the formula of the combination. In this question first, expand the summation and use the combination formulas ${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$ and ${}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}$ to calculate the value of summation.