Question

Choose the correct value of the binomial expression $$\sum\limits_{r = 0}^m {^{n + r}{c_n}} $$ from the options given below.
1) $^{n + m + 1}{c_{n + 1}}$
2) $^{n + m + 2}{c_n}$
3) $^{n + m + 3}{c_{n - 1}}$
4) None of these

Answer 1

Hint: Given a binomial expression to solve. We solve the given expression by using the binomial formula. The formulae used to solve the given expression are $^n{c_r}{ + ^n}{c_{r + 1}}{ = ^{n + 1}}{c_{r + 1}}$ ,$^n{c_0}{ = ^{n + 1}}{c_0}$, and $^n{c_r}{ = ^n}{c_{n - r}}$ . Using these formulas we solve the given expression step by step.

Complete step-by-step solution:
First of all, let’s change the given binomial expression in a convenient form
$$\sum\limits_{r = 0}^m {^{n + r}{c_n}} = \sum\limits_{r = 0}^m {^{n + r}{c_r}} $$
The above change can be done by using the formula $^n{c_r}{ = ^n}{c_{n - r}}$ .
Now, let’s continue expanding the summation
$$ \Rightarrow \sum\limits_{r = 0}^m {^{n + r}{c_n}} { = ^n}{c_0}{ + ^{n + 1}}{c_1}{ + ^{n + 2}}{c_2} + .....{ + ^{n + m}}{c_m}$$
Now, let’s use the formula $^n{c_0}{ = ^{n + 1}}{c_0}$ in the above equation. We get,
$$ \Rightarrow \sum\limits_{r = 0}^m {^{n + r}{c_n}} { = ^{n + 1}}{c_0}{ + ^{n + 1}}{c_1}{ + ^{n + 2}}{c_2} + .....{ + ^{n + m}}{c_m}$$
Now by using the formula $^n{c_r}{ + ^n}{c_{r + 1}}{ = ^{n + 1}}{c_{r + 1}}$ . We get,
$$ \Rightarrow \sum\limits_{r = 0}^m {^{n + r}{c_n}} { = ^{n + 2}}{c_1}{ + ^{n + 2}}{c_2}{ + ^{n + 3}}{c_2} + .....{ + ^{n + m}}{c_m}$$
By using the formula $^n{c_r}{ + ^n}{c_{r + 1}}{ = ^{n + 1}}{c_{r + 1}}$ to the above equation m-1 times we get,
$$ \Rightarrow \sum\limits_{r = 0}^m {^{n + r}{c_n}} { = ^{n + m + 1}}{c_m}$$
Now by using the formula $^n{c_r}{ = ^n}{c_{n - r}}$ . We get,
$$ \Rightarrow \sum\limits_{r = 0}^m {^{n + r}{c_n}} { = ^{n + m + 1}}{c_{n + 1}}$$
So, the value of the given expression is equal to.$$^{n + m + 1}{c_{n + 1}}$$
The correct option is 1.
Additional Information: We used a formula $^n{c_r}{ + ^n}{c_{r + 1}}{ = ^{n + 1}}{c_{r + 1}}$ . Let’s check the proof of this equation
$$ \Rightarrow LHS = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} + \dfrac{{n!}}{{(r + 1)!\left( {n - r - 1} \right)!}}$$
Let’s take the common terms out
$$ \Rightarrow LHS = \dfrac{{n!}}{{r!(n - r - 1)!}}\{ \dfrac{1}{{(n - r)}} + \dfrac{1}{{r + 1}}\} $$
After some computation in curly bracts, we get
$$ \Rightarrow LHS = \dfrac{{n!}}{{r!(n - r - 1)!}}\{ \dfrac{{r + 1 + n - r}}{{(r + 1)(n - r)}}\} $$
On further simplification, we get
$$ \Rightarrow LHS = \dfrac{{n!}}{{r!(n - r - 1)!}}\{ \dfrac{{n + 1}}{{(r + 1)(n - r}}\} $$
$$ \Rightarrow LHS = \dfrac{{(n + 1)!}}{{(r + 1)!(n + 1 - r - 1)!}}$$
It is nothing but
$$ \Rightarrow LHS = {}^{n + 1}{c_{r + 1}}$$
$$ \Rightarrow LHS = RHS$$

Note: We have to be careful while computing the step $$\sum\limits_{r = 0}^m {^{n + r}{c_n}} { = ^{n + 2}}{c_1}{ + ^{n + 2}}{c_2}{ + ^{n + 3}}{c_2} + .....{ + ^{n + m}}{c_m}$$ . Here $$^n{c_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$$, where $n!$ is defined as the product of first n natural numbers. There is another form of answer for the given expression nothing but $$^{n + m + 1}{c_m}$$. $$^{n + m + 1}{c_m}$$ this is also a correct answer for the given expression since there is no option for $$^{n + m + 1}{c_m}$$ so we used another form of this expression that is $$^{n + m + 1}{c_{n + 1}}$$.