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Question: Find the value \(\sum\limits_{r = 1}^n {r\left( {{}^n{C_r} - {}^n{C_{r - 1}}} \right)} \) . A) \({...

Find the value ∑r=1nr(nCr−nCr−1)\sum\limits_{r = 1}^n {r\left( {{}^n{C_r} - {}^n{C_{r - 1}}} \right)} .
A) 2n+n+1{2^n} + n + 1
B) 2n−n+1{2^n} - n + 1
C) n−2n+1n - {2^n} + 1
D) n−2n−1n - {2^n} - 1

Explanation

Solution

The given expression is the difference of two combination formulas so we will expand the given formula and then use the summation formulae to simplify the given expression further. Make the necessary arrangements in the equations obtained to get to the final result.

Complete step-by-step answer:
The given expression is ∑r=1nr(nCr−nCr−1)\sum\limits_{r = 1}^n {r\left( {{}^n{C_r} - {}^n{C_{r - 1}}} \right)} .
We know that the formula for combination of rr objects out of nn objects is given as follows:
nCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}
Using the above formula, the given expression can be rewritten as:
⇒\Rightarrow ∑r=1nr(nCr−nCr−1)=∑r=1nr(n!(n−r)!r!−n!(n−r+1)!(r−1)!)\sum\limits_{r = 1}^n {r\left( {{}^n{C_r} - {}^n{C_{r - 1}}} \right)} = \sum\limits_{r = 1}^n {r\left( {\dfrac{{n!}}{{\left( {n - r} \right)!r!}} - \dfrac{{n!}}{{\left( {n - r + 1} \right)!\left( {r - 1} \right)!}}} \right)}
We also know that the property of factorial states that n!=(n−1)!nn! = \left( {n - 1} \right)!n .
Therefore, the above expression becomes,
⇒\Rightarrow ∑r=1nr(n!(n−r)!r!−n!(n−r+1)!(r−1)!)=∑r=1nn!r(n−r)!(r−1)!(1r−1n−r+1)\sum\limits_{r = 1}^n {r\left( {\dfrac{{n!}}{{\left( {n - r} \right)!r!}} - \dfrac{{n!}}{{\left( {n - r + 1} \right)!\left( {r - 1} \right)!}}} \right)} = \sum\limits_{r = 1}^n {\dfrac{{n!r}}{{\left( {n - r} \right)!\left( {r - 1} \right)!}}\left( {\dfrac{1}{r} - \dfrac{1}{{n - r + 1}}} \right)}
Simplify the above terms as follows:
⇒\Rightarrow ∑r=1nn!r(n−r)!(r−1)!(1r−1n−r+1)=∑r=1n(n+1)!(n−r+1)!(r−1)!\sum\limits_{r = 1}^n {\dfrac{{n!r}}{{\left( {n - r} \right)!\left( {r - 1} \right)!}}\left( {\dfrac{1}{r} - \dfrac{1}{{n - r + 1}}} \right)} = \sum\limits_{r = 1}^n {\dfrac{{\left( {n + 1} \right)!}}{{\left( {n - r + 1} \right)!\left( {r - 1} \right)!}}}
Now with respect to summation (n+1)!\left( {n + 1} \right)! is constant.
⇒\Rightarrow ∑r=1n(n+1)!(n−r+1)!(r−1)!=(n+1)!∑r=1n1(n−r+1)!(r−1)!\sum\limits_{r = 1}^n {\dfrac{{\left( {n + 1} \right)!}}{{\left( {n - r + 1} \right)!\left( {r - 1} \right)!}}} = \left( {n + 1} \right)!\sum\limits_{r = 1}^n {\dfrac{1}{{\left( {n - r + 1} \right)!\left( {r - 1} \right)!}}} … (1)
Now we will expand the summation, keep aside the term (n+1)!\left( {n + 1} \right)!.
Now first we will put n=1n = 1 .
Then the expression (1) becomes:
⇒\Rightarrow (n+1)!∑r=1n1(n−r+1)!(r−1)!=2\left( {n + 1} \right)!\sum\limits_{r = 1}^n {\dfrac{1}{{\left( {n - r + 1} \right)!\left( {r - 1} \right)!}}} = 2
Now we will put n=2n = 2 .
Then the expression (1) becomes:
⇒\Rightarrow (n+1)!∑r=1n1(n−r+1)!(r−1)!=3\left( {n + 1} \right)!\sum\limits_{r = 1}^n {\dfrac{1}{{\left( {n - r + 1} \right)!\left( {r - 1} \right)!}}} = 3
Now put n=3n = 3 .
Then the expression (1) becomes:
⇒\Rightarrow (n+1)!∑r=1n1(n−r+1)!(r−1)!=6\left( {n + 1} \right)!\sum\limits_{r = 1}^n {\dfrac{1}{{\left( {n - r + 1} \right)!\left( {r - 1} \right)!}}} = 6
Similarly, if we keep substituting different values of nn then we obtain the values of the expression (1) which are the same for the expression 2n−n+1{2^n} - n + 1.
None of the other options gives similar values.
Therefore, we conclude that ∑r=1nr(nCr−nCr−1)=2n−n+1\sum\limits_{r = 1}^n {r\left( {{}^n{C_r} - {}^n{C_{r - 1}}} \right)} = {2^n} - n + 1

Thus, the correct answer is B.

Note: We did not actually solve this problem but we simplified it to the stage where we can verify the answer from the given options. We can also actually expand the summation and then solve the equation further till we get the same expansion as the answer.