Cần gấp

N

nguyenbahiep1

$2C_{n}^{0}+ \frac{2^{2}}{2}C_{n}^{1}+ \frac{2^{3}}{3}C_{n}^{2}+...+ \frac{2^{n+1}}{n+1}C_{n}^{n}= \frac{6560}{n+1}$

em có thể làm theo hướng sau

xét khai triển

[laTEX](1+x)^n = C_n^0 + C_n^1.x + C_n^2.x^2 +....+C_n^{n-1}x^{n-1}+C_n^n.x^n [/laTEX]

nguyên hàm 2 vế

[laTEX]\frac{(1+x)^{n+1}}{n+1} = C_n^0.x + C_n^1.\frac{x^2}{2} + C_n^2.\frac{x^3}{3} +....+C_n^{n-1}\frac{x^{n}}{n}+C_n^n.\frac{x^{n+1}}{n+1}[/laTEX]

thay x = 2

[laTEX]\frac{(1+2)^{n+1}}{n+1} = C_n^0.2 + C_n^1.\frac{2^2}{2} + C_n^2.\frac{2^3}{3} +....+C_n^{n-1}\frac{2^{n}}{n}+C_n^n.\frac{2^{n+1}}{n+1}[/laTEX]


[laTEX]\Rightarrow \frac{3^{n+1}}{n+1} = \frac{6561}{n+1} \Rightarrow n = ?[/laTEX]
 
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