Đạo Hàm

Question

tìm số nguyên dương

$n$ biết

$1.2^0C^{1}_{2n+1}-2.2^1C^{2}_{2n+1}+3.2^2C^3_{2n+1}-...+(2n+1).2^nC^{2n+1}_{2n+1}=2005$

score 1 · Answer 1

Xét

$f(x) = -(1-x)^{2n+1}$

$f(x) = -\left(\sum_{i=0}^{2n+1}C_{2n+1}^{i}(-1)^ix^i\right)$

$f'(x) = -\left(\sum_{i=1}^{2n+1}C_{2n+1}^{i}(-1)^iix^{i-1}\right)$

$f'(x) = 1.x^0C_{2n+1}^1-2.x^1C_{2n+1}^2+3x^2C_{2n+1}^3 +....+(-1)^iix^{i-1}C_{2n+1}^i+...+(2n+1)x^{2n}C_{2n+1}^{2n+1}$

thay x = 2 ta được

$f'(2)=1.2^0C_{2n+1}^1-2.2^1C_{2n+1}^2+32^2C_{2n+1}^3 +....+(-1)^iix^{i-1}C_{2n+1}^i+...+(2n+1)2^{2n}C_{2n+1}^{2n+1}$

Mặt khác

$f'(x) = (2n+1)(1-x)^{2n}$

nên

$f'(2) = (2n+1)(1-2)^{2n} = (2n+1)$

Vậy

$1.2^0C_{2n+1}^1-2.2^1C_{2n+1}^2+32^2C_{2n+1}^3 +....+(-1)^iix^{i-1}C_{2n+1}^i+...+(2n+1)2^{2n}C_{2n+1}^{2n+1} = (2n+1) =2005$

suy ra

$n =1002$

Nhớ vote nhé

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