First slide

Binomial theorem for positive integral Index

Question

$(\begin{matrix} 30 \\ 0 \end{matrix}) (\begin{array}{l} 30 \\ 10 \end{array}) - (\begin{matrix} 30 \\ 1 \end{matrix}) (\begin{array}{l} 30 \\ 11 \end{array}) + \dots + (\begin{array}{l} 30 \\ 20 \end{array}) (\begin{array}{l} 30 \\ 30 \end{array})$ is equal to

Moderate

A

$^{30} C_{11}$
B

$^{60} C_{10}$
C

$^{30} C_{10}$
D

$^{65} C_{55}$

Solution

$Let A = (\begin{matrix} 30 \\ 0 \end{matrix}) (\begin{array}{l} 30 \\ 10 \end{array}) - (\begin{matrix} 30 \\ 1 \end{matrix}) (\begin{array}{l} 30 \\ 11 \end{array}) + (\begin{matrix} 30 \\ 2 \end{matrix}) (\begin{array}{l} 30 \\ 12 \end{array}) - \dots + (\begin{array}{l} 30 \\ 20 \end{array}) (\begin{array}{l} 30 \\ 30 \end{array})$
$or \begin{array}{l} A =^{30} C_{0} \cdot^{30} C_{10} -^{30} C_{1} \cdot^{30} C_{11} +^{30} C_{2} \cdot^{30} C_{12} - \dots +^{30} C_{20} \cdot^{30} C_{30} \end{array}$
$\begin{array}{l} = Coefficient of x^{20} in (1 + x)^{30} (1 - x)^{30} \\ = Coefficient of x^{20} in {(1 - x^{2})}^{30} \\ = Coefficient of x^{20} in \sum_{r = 0}^{30} {(- 1)^{r}}^{30} C_{r} {(x^{2})}^{r} \\ = {(- 1)^{10}}^{30} C_{10} [for coefficient of x^{20}, let r = 10] \\ =^{30} C_{10} \end{array}$
$\begin{array}{l} = (- 1)^{10 30} C_{10} [for coefficient of x^{20}, let r = 10] \\ =^{30} C_{10} \end{array}$

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