First slide

Binomial theorem for positive integral Index

Question

$If f (m) = \sum_{i = 0}^{m} (\begin{matrix} 30 \\ 30 - i \end{matrix}) (\begin{matrix} 20 \\ m - i \end{matrix}) where (\begin{array}{l} p \\ q \end{array}) =^{p} C_{q,} then$

Difficult

A

$maximum value of f (m) {is}^{50} C_{25}$
B

$f (0) + f (1) + \dots + f (50) = 2^{50}$
C

$f (m) is always divisible by 50 (1 \leq m \leq 49)$
D

$The value of \sum_{m = 0}^{51} (f (m))^{2} =^{100} C_{50}$

Solution

$f (m) = \sum_{i = 0}^{m} (\begin{matrix} 30 \\ 30 - i \end{matrix}) (\begin{matrix} 20 \\ m - i \end{matrix}) = \sum_{i = 0}^{m} (\begin{matrix} 30 \\ i \end{matrix}) (\begin{matrix} 20 \\ m - i \end{matrix}) =^{50} C_{m}$
$f (m) is greatest when m = 25 . Also,$
$\begin{array}{l} f (0) + f (1) + \dots + f (50) \\ =^{50} C_{0} +^{50} C_{1} +^{50} C_{2} + \dots +^{50} C_{50} = 2^{50} \end{array}$
${Also,}^{50} C_{m} is not divisible by 50 for any m as 50 is not a prime number$
$\sum_{m = 0}^{50} (f (m))^{2} = {(^{50} C_{0})}^{2} + {(^{50} C_{1})}^{2} + {(^{50} C_{2})}^{2} + \dots + {(^{50} C_{50})}^{2} =^{100} C_{50}$

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