First slide

Binomial theorem for positive integral Index

Question

For $0 \leq r < 2 n, (^{2 n + r} C_{n}) (^{2 n - r} C_{n})$ cannot exceed

Moderate

A

$^{4 n} C_{n}$
B

$^{4 n} C_{2 n}$
C

$^{6 n} C_{3 n}$
D

none of these.

Solution

For $0 \leq r < 2 n,$
$\begin{array}{l} (1 + x)^{4 n} = (1 + x)^{2 n + r} (1 + x)^{2 n - r} \\ = (A_{0} + A_{1} x + A_{2} x^{2} + \dots + A_{2 n + r} x^{2 n + r}) (B_{0} + B_{1} x + B_{2} x^{2} + \dots + B_{2 n - r} x^{2 n - r}) \end{array}$
where $A_{k} =^{2 n + r} C_{k} (0 \leq k \leq 2 n + r)$
and $B_{k} =^{2 n - r} C_{k} (0 \leq k \leq 2 n - r)$
Coefficient of $x^{2 n}$ on the RHS
$\begin{array}{l} A_{2 n} B_{0} + A_{2 n - 1} B_{1} + \dots + A_{n} B_{n} + A_{n + 1} B_{n - 1} + \dots + A_{r} B_{2 n - r} \end{array}$
= coefficient of $x^{2 n}$ on LHS
$=^{4 n} C_{2 n} .$
Thus, $A_{n} B_{n} <^{4 n} C_{2 n}$
$\Rightarrow (^{2 n + r} C_{n}) (^{2 n - r} C_{n}) <^{4 n} C_{2 n}$

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