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

Moderate

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

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,
$f (0) + f (1) + \dots + f (50) =^{50} C_{0} +^{50} C_{1} +^{50} C_{2} + \dots +^{50} C_{50} = 2^{50}$
Also, $^{50} C_{m}$ s not divisible by 50 for all m as 50 is not a prime number
$\begin{matrix} \sum_{m = 0}^{50} (f (m))^{2} = {(^{50} C_{0})}^{2} + {(^{50} C_{1})}^{2} + {(^{50} C_{2})}^{2} + \dots + {(^{50} C_{50})}^{2} \\ =^{100} C_{50} \end{matrix}$

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