$C_{0} C_{2} + C_{1} C_{3} + C_{2} C_{4} + . . . + C_{n - 2} C_{n} =^{2 n} C_{n - 1}$ or $^{2 n} C_{n + 1}$

True

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False

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Solution

The correct option is B False
From the product of (i) and (ii) comparing coefficients of $x^{n - 2}$ or $x^{n + 2}$ both sides,
$C_{0} C_{2} + C_{1} C_{3} + C_{2} C_{4} + . . . + C_{n - 2} C_{n} =^{2 n} C_{n - 2}$ or $^{2 n} C_{n + 2} .$

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Similar questions

C_{0} C_{2} + C_{1} C_{3} + C_{2} C_{4} + C_{n - 2} C_{n} = \frac{(2 n)!}{(n - 1)! (n = 1)!}

(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . + C_{n} x^{n}

C_{0} C_{2} + C_{1} C_{3} + C_{2} C_{4} + . . . + C_{n - 2} C_{n} =

C_{0} C_{2} + C_{1} C_{3} + C_{2} C_{4} + . . . . . + C_{n - 2} C_{n},

(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + C_{3} x^{3} + \dots + C_{n} x^{n},

C_{0} C_{2} + C_{1} C_{3} + C_{2} C_{4} + \dots + C_{n - 2} C_{n} =

\frac{1}{2}

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