If $n$ is an integer greater than $1$, then $a{ - ^n}{C_1}(a - 1){ + ^n}{C_2}(a - 2) + .... + {( - 1)^n}(a - n) = $

  • [IIT 1972]
  • A

    $a$

  • B

    $0$

  • C

    ${a^2}$

  • D

    ${2^n}$

Similar Questions

Let ${s_1} = \mathop \sum \limits_{j = 1}^{10} j\left( {j - 1} \right)\left( {\begin{array}{*{20}{c}}{10}\\j\end{array}} \right)\;,$$\;{s_2} = \mathop \sum \limits_{j = 1}^{10} j\;\left( {\begin{array}{*{20}{c}}{10}\\j\end{array}} \right)\;and,$${s_3} = \mathop \sum \limits_{j = 1}^{10} {j^2}\left( {\begin{array}{*{20}{c}}{10}\\j\end{array}} \right)\;,\;$

Statement $-1$:${s_3} = 55 \times {2^9}$

Statement $-2$: ${s_1} = 90 \times {2^8}\;$ and ${s_2} = 10 \times {2^8}$ 

  • [AIEEE 2010]

If ${S_n} = \sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}}} $ and ${t_n} = \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} $, then $\frac{{{t_n}}}{{{S_n}}}$ is equal to

  • [AIEEE 2004]

$\left( {\left( {\begin{array}{*{20}{c}}
{21}\\
1
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
1
\end{array}} \right)} \right) + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
2
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
2
\end{array}} \right)} \right)$$ + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
3
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
3
\end{array}} \right)} \right) + \;.\;.\;.$$ + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
{10}
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
{10}
\end{array}} \right)} \right) = $

  • [JEE MAIN 2017]

If $\sum\limits_{ k =1}^{31}\left({ }^{31} C _{ k }\right)\left({ }^{31} C _{ k -1}\right)-\sum\limits_{ k =1}^{30}\left({ }^{30} C _{ k }\right)\left({ }^{30} C _{ k -1}\right)=\frac{\alpha(60 !)}{(30 !)(31 !)}$

Where $\alpha \in R$, then the value of $16 \alpha$ is equal to

  • [JEE MAIN 2022]

The value of $\frac{{{C_1}}}{2} + \frac{{{C_3}}}{4} + \frac{{{C_5}}}{6} + .....$ is equal to