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Combination

Prove that, ...

Question

Prove that, $(^{2 n} C_{1})^{2} + 2. (^{2 n} C_{1})^{2} + 3. (^{2 n} C_{3})^{2} + . . . . + 2 n . (^{2 n} C_{2 n})^{2} = \frac{(4 n - 1)!}{{(2 n - 1)!}^{2}}$ .

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Solution

$(1 + x)^{2 n} =^{2 n} C_{0} +^{2 n} C_{1} x +^{2 n} C_{2} x^{2} + . . . . +^{2 n} C_{2 n} x^{2 n} - - - - - (1)$
Differentiating eq (1) w.r. to $x$

$2 n (1 + x)^{2 n - 1} =^{2 n} C_{1} + 2^{2 n} C_{2} x + . . . . + 2 n \cdot^{2 n} C_{2 n} x^{2 n - 1} - - - - - (2)$

We know that
${(1 + \frac{1}{x})}^{2 n} =^{2 n} C_{0} + \frac{^{2 n} C_{1}}{x} + \frac{^{2 n} C_{2}}{x^{2}} + . . . . + \frac{^{2 n} C_{2 n}}{x^{2 n}} - - - - - (3)$

Multiplying eq (2) and (3) and comparing $x^{r}$ we get
$^{2 n} C_{1}^{2 n} C_{r + 1} + 2 \cdot^{2 n} C_{2}^{2 n} C_{r + 2} + . . . . . . + 2 n \cdot^{2 n} C_{2 n}^{2 n - r} C_{2 n} = 2 n \cdot^{4 n - 1} C_{2 n + r}$

Putting $r = 0$
$(^{2 n} C_{1})^{2} + 2 \cdot (^{2 n} C_{2})^{2} + . . . . . . + 2 n \cdot (^{2 n} C_{2 n})^{2} C_{2 n} = 2 n \cdot^{4 n - 1} C_{2 n}$

$(^{2 n} C_{1})^{2} + 2 \cdot (^{2 n} C_{2})^{2} + . . . . . . + 2 n \cdot (^{2 n} C_{2 n})^{2} C_{2 n} = 2 n \cdot \frac{(4 n - 1)!}{(2 n)! (4 n - 1 - 2 n)!}$

$(^{2 n} C_{1})^{2} + 2 \cdot (^{2 n} C_{2})^{2} + . . . . . . + 2 n \cdot (^{2 n} C_{2 n})^{2} C_{2 n} = \cdot \frac{(4 n - 1)!}{(2 n - 1)! (2 n - 1)!}$

$(^{2 n} C_{1})^{2} + 2 \cdot (^{2 n} C_{2})^{2} + . . . . . . + 2 n \cdot (^{2 n} C_{2 n})^{2} C_{2 n} = \cdot \frac{(4 n - 1)!}{((2 n - 1)!)^{2}}$

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Similar questions

\frac{^{4 n} C_{2 n}}{^{2 n} C_{n}} = \frac{1 \cdot 3 \cdot 5 . . . . (4 n - 1)}{[1 \cdot 3 \cdot 5 . . . . (2 n - 1)]^{2}}

1.3 + 3.5 + 5.7 + . . . . + (2 n - 2 n + 1) = \frac{n (4 n^{2} + 6 n - 1)}{3}

\frac{^{4 n} C_{2 n}}{^{2 n} C_{n}} = \frac{1.3.5...... (4 n - 1)}{{1.3.5...... (2 n - 1)}^{2}}

^{4 n} C_{2 n} :^{2 n} C_{n} = 1.3.5... (4 n - 1) : [1.3.5... (2 n - 1)]^{2} .

Q. Prove that: ⁴ⁿC_2n : ²ⁿC_n = [1 · 3 · 5 ... (4n − 1)] : [1 · 3 · 5 ... (2n − 1)]².

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