prove that $C_{0}^{2 n} C_{n} -^{2 n - 2} C_{n} +^{2 n - 4} C_{n} = 2^{n} w h e r e C_{r} =^{n} C_{r}$

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Solution

Consider the expansion of
$(2 x + x^{2})^{n} = [(1 + x)^{2} - 1]^{n}$
or $x^{n} (2 + x)^{n} = [(1 + x)^{2} - 1]^{n}$
The coefficient of $x^{n}$ in L. H. S. is $2^{n}$ and it should be equal to coefficient of $x^{n}$ in R. H. S.
$[(1 + x)^{2} - 1]^{n} = C_{0} (1 + x) 2 n - C_{1} (1 + x) 2 n - 2 + C_{2} (1 + x) 2 n - 4$
$- = C_{0}^{2 n} C_{n} - C_{1}^{2 n - 2} C_{n} + C_{2}^{2 n - 4} C_{n}$
We konw that the coefficient of $s^{r} i n (1 + x)^{n} i s^{n} C_{r}$

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2^{2} \frac{C_{0}}{1.2} + 2^{3} \frac{C_{1}}{2.3} + 2^{4} \frac{C_{2}}{3.4} + . . . . . . . . . + 2^{n + 2} \frac{C_{n}}{(n + 1) (n + 2)} = \frac{3^{n + 2} - 2 n - 5}{(n + 1) (n + 2)}

∣ ∣ ∣ ∣ ∣ \begin{matrix} {n + 2}_{C_{n}} & n + 3_{C_{n + 1}} & n + 4_{C_{n + 2}} n + 3_{C_{n + 1}} & n + 4_{C_{n + 2}} & n + 5_{C_{n + 3}} n + 4_{C_{n + 2}} & n + 5_{C_{n + 3}} & n + 6_{C_{n + 4}} \end{matrix} ∣ ∣ ∣ ∣ ∣

(1 + x)^{n} = \sum_{r = 0}^{n} C_{r} x^{r}

\frac{2^{2} C_{0}}{1.2} + \frac{2^{3} C_{1}}{2.3} + . . . . . . + \frac{2^{n + 2} C_{n}}{(n + 1) (n + 2)}

n

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c_{r}

^{n} C_{r}

C_{1} s i n α c o s (n - 1) α + C_{2} s i n 2 α c o s (n - 2) α + C_{3} s i n 3 α c o s (n - 3) α + . . + C_{n} s i n n α

= 2^{n - 1} s i n n α

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