Cho Khai Triển ${{\left( 1+2x \right)}^{2014}}={{a}_{0}}+{{a}_{1}}x+{{a}

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You are using an out of date browser. It may not display this or other websites correctly.You should upgrade or use an alternative browser. Cho khai triển ${{\left( 1+2x \right)}^{2014}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+…+{{a}_{2014}}{{x}^{2014}}$. Khi đó tổng $S={{a}_{1}}+{{3

• Thread starter Julipkq
• Start date Jun 21, 2021

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Julipkq

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Cho khai triển ${{\left( 1+2x \right)}^{2014}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+…+{{a}_{2014}}{{x}^{2014}}$. Khi đó tổng $S={{a}_{1}}+{{3}^{2}}{{a}_{3}}+…+{{3}^{2010}}{{a}_{2011}}+{{3}^{2012}}{{a}_{2013}}$ có giá trị bằng C. $\frac{{{7}^{2014}}-{{5}^{2014}}}{6}$. B. $\frac{{{7}^{2014}}-{{5}^{2014}}}{2}$. C. $\frac{{{7}^{2014}}}{6}$. D. $\frac{{{5}^{2014}}}{2}$. Sort by date Sort by votes N

Nguyễn Thu Trang

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Đáp án A. Nhận thấy rằng: $3S=3{{a}_{1}}+{{3}^{3}}{{a}_{3}}+…+{{3}^{2011}}{{a}_{2011}}+{{3}^{2013}}{{a}_{2013}}$ Lần lượt thay $x=3$,$x=-3$ vào khai triển đã cho ta được: $P\left( 3 \right)={{7}^{2014}}={{a}_{0}}+3{{a}_{1}}+{{3}^{2}}{{a}_{2}}+…+{{3}^{2013}}{{a}_{2013}}+{{3}^{2014}}{{a}_{2014}}$ $P\left( -3 \right)={{5}^{2014}}={{a}_{0}}-3{{a}_{1}}+{{3}^{2}}{{a}_{2}}-…-{{3}^{2013}}{{a}_{2013}}+{{3}^{2014}}{{a}_{2014}}$ Trừ hai đẳng thức này vế theo vế, ta được: $2\left( 3{{a}_{1}}+{{3}^{3}}{{a}_{3}}…+{{3}^{2011}}{{a}_{2011}}+{{3}^{2013}}{{a}_{2013}} \right)={{7}^{2014}}-{{5}^{2014}}$ $\Leftrightarrow 3S=\frac{{{7}^{2014}}-{{5}^{2014}}}{2}\Leftrightarrow S=\frac{{{7}^{2014}}-{{5}^{2014}}}{6}$ Vậy $S={{a}_{1}}+{{3}^{2}}{{a}_{3}}+…+{{3}^{2010}}{{a}_{2011}}+{{3}^{2012}}{{a}_{2013}}=\frac{{{7}^{2014}}-{{5}^{2014}}}{6}$ Upvote 0 Downvote You must log in or register to reply here.

Tìm $n$ sao cho $C_{2n}^{1}+C_{2n}^{3}+…C_{2n}^{2n-1}=2048.$ là C. $n=8$. B. $n=6$. C. $n=7$. D...

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