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Binomial Coefficients

If c0,c1,c2...

Question

If $c_{0}, c_{1}, c_{2}, . . . . . . . c_{n}$ denote the coefficients in the expansion of ${(1 + x)}^{n}$ , prove that
$c_{0} c_{r} + c_{1} c_{r + 1} + c_{2} c_{r + 2} + . . . . + c_{n - r} c_{n} = \frac{| 2 n - - -}{| n - r - ---- - | n + r - ---- -}$ .

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Solution

$⟹ {(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x + . . . . . . . + C_{n} \cdot x^{n} ⟶ (I)$
${(x + 1)}^{n} = C_{0} x^{n} + C_{1} x^{n - 1} + . . . . . . . C_{r} x^{n - r} + C_{r + 1} x^{n - r + 1} + . . . . . + C_{n} x^{0} ⟶ (I I)$
Multiplying $(I)$ and $(I I)$ , we get
$C_{0} C_{r} + C_{1} C_{r + 1} + C_{2} C_{r + 2} + . . . . . + C_{n - r} C_{r}$
$=$ coefficient of $x^{n - r}$ in ${(1 + x)}^{2 n}$
$=^{2 n} C_{n - r}$
$= \frac{(2 n)!}{(n - r)! (n + r)!}$

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Similar questions

c_{0}, c_{1}, c_{2}, . . . . . . . c_{n}

denote the coefficients in the expansion of

{(1 + x)}^{n}

(c_{0} + c_{1}) (c_{1} + c_{2}) . . . . . . . (c_{n - 1} + c_{n}) = \frac{c_{1} c_{2} . . . c_{n} {(n + 1)}^{n}}{| n - -}

c_{0}, c_{1}, c_{2}, . . . . . . . c_{n}

denote the coefficients in the expansion of

{(1 + x)}^{n}

c_{0} + \frac{c_{1}}{2} + \frac{c_{2}}{3} + . . . . . . + \frac{c_{n}}{n + 1} = \frac{2^{n + 1} - 1}{n + 1}

c_{0}, c_{1}, c_{2}, . . . . c_{n}

denote the coefficients in the expansion of

(1 + x)^{n}

c_{0} + \frac{c_{1}}{2} + \frac{c_{2}}{3} + . . . . . + \frac{c_{n}}{n + 1} = \frac{2^{n + 1} - 1}{n + 1}

r = 0, 1, 2, . . . ., n

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

and hence deduce that
i)

C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + . . . . . . + C_{n}^{2} =^{2 n} C_{n}

C_{0} \cdot C_{1} + C_{1} \cdot C_{2} + C_{2} \cdot C_{3} + . . . . . + C_{n - 1} \cdot C_{n} =^{2 n} C_{n + 1}

c_{0}, c_{1}, c_{2}, . . . . c_{n}

denote the coefficients in the expansion of

(1 + x)^{n}

\frac{c_{1}}{c_{0}} + \frac{2 c_{2}}{c_{1}} + \frac{3 c_{3}}{c_{2}} + . . . . . + \frac{n c_{n}}{c_{n - 1}} = \frac{n (n + 1)}{2}

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