Solveeit Logo
Login

Question

Question: Choose the correct value of the binomial expression \[\sum\limits_{r = 0}^m {^{n + r}{c_n}} \] from ...

Choose the correct value of the binomial expression ∑r=0mn+rcn\sum\limits_{r = 0}^m {^{n + r}{c_n}} from the options given below.

  1. n+m+1cn+1^{n + m + 1}{c_{n + 1}}
  2. n+m+2cn^{n + m + 2}{c_n}
  3. n+m+3cn−1^{n + m + 3}{c_{n - 1}}
  4. None of these
Explanation

Solution

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 ncr+ncr+1=n+1cr+1^n{c_r}{ + ^n}{c_{r + 1}}{ = ^{n + 1}}{c_{r + 1}} ,nc0=n+1c0^n{c_0}{ = ^{n + 1}}{c_0}, and ncr=ncn−r^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
∑r=0mn+rcn=∑r=0mn+rcr\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 ncr=ncn−r^n{c_r}{ = ^n}{c_{n - r}} .
Now, let’s continue expanding the summation
⇒∑r=0mn+rcn=nc0+n+1c1+n+2c2+.....+n+mcm\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 nc0=n+1c0^n{c_0}{ = ^{n + 1}}{c_0} in the above equation. We get,
⇒∑r=0mn+rcn=n+1c0+n+1c1+n+2c2+.....+n+mcm\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 ncr+ncr+1=n+1cr+1^n{c_r}{ + ^n}{c_{r + 1}}{ = ^{n + 1}}{c_{r + 1}} . We get,
⇒∑r=0mn+rcn=n+2c1+n+2c2+n+3c2+.....+n+mcm\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 ncr+ncr+1=n+1cr+1^n{c_r}{ + ^n}{c_{r + 1}}{ = ^{n + 1}}{c_{r + 1}} to the above equation m-1 times we get,
⇒∑r=0mn+rcn=n+m+1cm\Rightarrow \sum\limits_{r = 0}^m {^{n + r}{c_n}} { = ^{n + m + 1}}{c_m}
Now by using the formula ncr=ncn−r^n{c_r}{ = ^n}{c_{n - r}} . We get,
⇒∑r=0mn+rcn=n+m+1cn+1\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+1cn+1^{n + m + 1}{c_{n + 1}}
The correct option is 1.
Additional Information: We used a formula ncr+ncr+1=n+1cr+1^n{c_r}{ + ^n}{c_{r + 1}}{ = ^{n + 1}}{c_{r + 1}} . Let’s check the proof of this equation
⇒LHS=n!r!(n−r)!+n!(r+1)!(n−r−1)!\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
⇒LHS=n!r!(n−r−1)!1(n−r)+1r+1\Rightarrow LHS = \dfrac{{n!}}{{r!(n - r - 1)!}}\\{ \dfrac{1}{{(n - r)}} + \dfrac{1}{{r + 1}}\\}
After some computation in curly bracts, we get
⇒LHS=n!r!(n−r−1)!r+1+n−r(r+1)(n−r)\Rightarrow LHS = \dfrac{{n!}}{{r!(n - r - 1)!}}\\{ \dfrac{{r + 1 + n - r}}{{(r + 1)(n - r)}}\\}
On further simplification, we get
⇒LHS=n!r!(n−r−1)!n+1(r+1)(n−r\Rightarrow LHS = \dfrac{{n!}}{{r!(n - r - 1)!}}\\{ \dfrac{{n + 1}}{{(r + 1)(n - r}}\\}
⇒LHS=(n+1)!(r+1)!(n+1−r−1)!\Rightarrow LHS = \dfrac{{(n + 1)!}}{{(r + 1)!(n + 1 - r - 1)!}}
It is nothing but
⇒LHS=n+1cr+1\Rightarrow LHS = {}^{n + 1}{c_{r + 1}}
⇒LHS=RHS\Rightarrow LHS = RHS

Note: We have to be careful while computing the step ∑r=0mn+rcn=n+2c1+n+2c2+n+3c2+.....+n+mcm\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 ncr=n!r!(n−r)!^n{c_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}, where n!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+1cm^{n + m + 1}{c_m}. n+m+1cm^{n + m + 1}{c_m} this is also a correct answer for the given expression since there is no option for n+m+1cm^{n + m + 1}{c_m} so we used another form of this expression that is n+m+1cn+1^{n + m + 1}{c_{n + 1}}.