# The value of $\sum\limits_{r = 1}^{10} {r.\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}}} $ is equal to

$

a.{\text{ 5}}\left( {2n - 9} \right) \\

b.{\text{ 10}}n \\

c.{\text{ 9}}\left( {n - 4} \right) \\

d.{\text{ }}\left( {n - 2} \right) \\

$

Last updated date: 27th Mar 2023

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Answer

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Hint – In this question use the property of combination which is given as ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ so, use this property to reach the answer.

Given equation is

$\sum\limits_{r = 1}^{10} {r.\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}}} $……………. (1)

As we know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ (combination property)

So, ${}^n{C_{r - 1}} = \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}}$

Now, we know that $r! = r\left( {r - 1} \right)!{\text{ \& }}\left( {n - r + 1} \right)! = \left( {n - r + 1} \right)\left( {n - r} \right)!$

So substitute these values in equation (1) and simplify we get

$\sum\limits_{r = 1}^{10} {r.\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}}} = \sum\limits_{r = 1}^{10} {r.\dfrac{{\dfrac{{n!}}{{r!\left( {n - r} \right)!}}}}{{\dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}}}}} = \sum\limits_{r = 1}^{10} {r.} \dfrac{{\dfrac{{n!}}{{r\left( {r - 1} \right)!\left( {n - r} \right)!}}}}{{\dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)\left( {n - r} \right)!}}}}$

Now simplify the above equation we have

$\sum\limits_{r = 1}^{10} {r.\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}}} = \sum\limits_{r = 1}^{10} {\left( {n - r + 1} \right)} = \sum\limits_{r = 1}^{10} {\left( {\left( {n + 1} \right) - r} \right)} = \sum\limits_{r = 1}^{10} {\left( {n + 1} \right) - \sum\limits_{r = 1}^{10} r } $

(n + 1) is constant w.r.t. r so it is written outside the summation therefore

$\sum\limits_{r = 1}^{10} {r.\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}}} = \left( {n + 1} \right)\sum\limits_{r = 1}^{10} {1 - \sum\limits_{r = 1}^{10} r } $

$\sum\limits_{r = 1}^{10} {r.\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}}} = \left( {n + 1} \right)\sum\limits_{r = 1}^{10} {1 - \sum\limits_{r = 1}^{10} r } .............\left( 2 \right)$

Now as we know that $\sum\limits_{r = 1}^n 1 = n{\text{, }}\sum\limits_{r = 1}^n r = \dfrac{{n\left( {n + 1} \right)}}{2}$

But in the above equation r is from 1 to 10.

$\sum\limits_{r = 1}^{10} 1 = 10{\text{, }}\sum\limits_{r = 1}^{10} r = \dfrac{{10\left( {10 + 1} \right)}}{2}$

Therefore from equation (2)

$

\sum\limits_{r = 1}^{10} {r.\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}}} = \left( {n + 1} \right)\sum\limits_{r = 1}^{10} {1 - \sum\limits_{r = 1}^{10} r } = \left( {n + 1} \right)10 - \dfrac{{10\left( {10 + 1} \right)}}{2} \\

\Rightarrow \sum\limits_{r = 1}^{10} {r.\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}}} = 10n + 10 - 55 = 10n - 45 = 5\left( {2n - 9} \right) \\

$

Hence, option (a) is correct.

Note – In such types of questions the key concept we have to remember is that always recall the property of combination and values of summation which is all stated above, then apply these properties in the given equation and simplify we will get the required answer.

Given equation is

$\sum\limits_{r = 1}^{10} {r.\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}}} $……………. (1)

As we know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ (combination property)

So, ${}^n{C_{r - 1}} = \dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}}$

Now, we know that $r! = r\left( {r - 1} \right)!{\text{ \& }}\left( {n - r + 1} \right)! = \left( {n - r + 1} \right)\left( {n - r} \right)!$

So substitute these values in equation (1) and simplify we get

$\sum\limits_{r = 1}^{10} {r.\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}}} = \sum\limits_{r = 1}^{10} {r.\dfrac{{\dfrac{{n!}}{{r!\left( {n - r} \right)!}}}}{{\dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}}}}} = \sum\limits_{r = 1}^{10} {r.} \dfrac{{\dfrac{{n!}}{{r\left( {r - 1} \right)!\left( {n - r} \right)!}}}}{{\dfrac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)\left( {n - r} \right)!}}}}$

Now simplify the above equation we have

$\sum\limits_{r = 1}^{10} {r.\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}}} = \sum\limits_{r = 1}^{10} {\left( {n - r + 1} \right)} = \sum\limits_{r = 1}^{10} {\left( {\left( {n + 1} \right) - r} \right)} = \sum\limits_{r = 1}^{10} {\left( {n + 1} \right) - \sum\limits_{r = 1}^{10} r } $

(n + 1) is constant w.r.t. r so it is written outside the summation therefore

$\sum\limits_{r = 1}^{10} {r.\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}}} = \left( {n + 1} \right)\sum\limits_{r = 1}^{10} {1 - \sum\limits_{r = 1}^{10} r } $

$\sum\limits_{r = 1}^{10} {r.\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}}} = \left( {n + 1} \right)\sum\limits_{r = 1}^{10} {1 - \sum\limits_{r = 1}^{10} r } .............\left( 2 \right)$

Now as we know that $\sum\limits_{r = 1}^n 1 = n{\text{, }}\sum\limits_{r = 1}^n r = \dfrac{{n\left( {n + 1} \right)}}{2}$

But in the above equation r is from 1 to 10.

$\sum\limits_{r = 1}^{10} 1 = 10{\text{, }}\sum\limits_{r = 1}^{10} r = \dfrac{{10\left( {10 + 1} \right)}}{2}$

Therefore from equation (2)

$

\sum\limits_{r = 1}^{10} {r.\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}}} = \left( {n + 1} \right)\sum\limits_{r = 1}^{10} {1 - \sum\limits_{r = 1}^{10} r } = \left( {n + 1} \right)10 - \dfrac{{10\left( {10 + 1} \right)}}{2} \\

\Rightarrow \sum\limits_{r = 1}^{10} {r.\dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}}} = 10n + 10 - 55 = 10n - 45 = 5\left( {2n - 9} \right) \\

$

Hence, option (a) is correct.

Note – In such types of questions the key concept we have to remember is that always recall the property of combination and values of summation which is all stated above, then apply these properties in the given equation and simplify we will get the required answer.

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