Question

Đã trả lời bởi chuyên gia

Answer 1

Hướng dẫn giải

a) $1 + 2 C_{n}^{1} + 4 C_{n}^{2} + \dots + 2^{n - 1} C_{n}^{n - 1} + 2^{n} C_{n}^{n}$

$= C_{n}^{0} 1 + C_{n}^{1} 2 + C_{n}^{2} 2^{2} + \dots + C_{n}^{n - 1} 2^{n - 1} + C_{n}^{n} 2^{n}$

$= C_{n}^{0} 1^{n} + C_{n}^{1} 1^{n - 1} 2 + C_{n}^{2} 1^{n - 2} 2^{2} + \dots + C_{n}^{n - 1} 1 {.2}^{n - 1} + C_{n}^{n} 2^{n}$

= (1 + 2)n = 3n.

b) Ta có:

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

= C_{2 n}^{0} x^{2 n} + C_{2 n}^{1} x^{2 n - 1} + C_{2 n}^{2} x^{2 n - 2} + \dots + C_{2 n}^{2 n - 1} x + C_{2 n}^{2 n} .

Cho x = –1, ta được:

${(- 1 + 1)}^{2 n} = C_{2 n}^{0} {(- 1)}^{2 n} {+C}_{2 n}^{1} {(- 1)}^{2 n - 1} + C_{2 n}^{2} {(- 1)}^{2 n - 2} + \dots + C_{2 n}^{2 n - 1} (- 1) + C_{2 n}^{2 n}$

$= C_{2 n}^{0} - C_{2 n}^{1} + C_{2 n}^{2} - \dots - C_{2 n}^{2 n - 1} + C_{2 n}^{2 n}$

\Rightarrow C_{2 n}^{0} - C_{2 n}^{1} + C_{2 n}^{2} - \dots - C_{2 n}^{2 n - 1} + C_{2 n}^{2 n} = 0

\Rightarrow C_{2 n}^{0} + C_{2 n}^{2} + C_{2 n}^{4} + \dots + C_{2 n}^{2 n} = C_{2 n}^{1} + C_{2 n}^{3} + C_{2 n}^{5} + \dots + C_{2 n}^{2 n - 1} .

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Answer 2