Tính: $S=2.C_{2020}^{1}+3.2^3.C_{2020}{3}+…+2019.2^2019.C_{2020}^{2019}$ 03/10/2021 Bởi Gabriella Tính: $S=2.C_{2020}^{1}+3.2^3.C_{2020}{3}+…+2019.2^2019.C_{2020}^{2019}$
Giải thích các bước giải: Ta có: $S = 2.C_{2020}^1 + {3.2^3}.C_{2020}^3 + … + {2019.2^{2019}}.C_{2020}^{2019}$ Số hạng tổng quát của dãy tổng $S$ là: $\begin{array}{l}k{.2^k}.C_{2020}^k\left( {1 \le k \le 2019;k\not \vdots 2} \right)\\ = {2^k}.k.C_{2020}^k\\ = {2^k}.2020.C_{2019}^{k – 1}\\ = 4040.C_{2019}^{k – 1}{.2^{k – 1}}\end{array}$ Khi đó: $S = 4040\left( {C_{2019}^0 + C_{2019}^2{{.2}^2} + C_{2019}^4{{.2}^4} + … + C_{2019}^{2018}{{.2}^{2018}}} \right)$ Ta xét khai triển: $\begin{array}{l} + ){\left( {1 + x} \right)^{2019}} = C_{2019}^0 + C_{2019}^1.x + C_{2019}^2.{x^2} + … + C_{2019}^{2018}.{x^{2018}} + C_{2019}^{2019}.{x^{2019}}\\ + ){\left( {1 – x} \right)^{2019}} = C_{2019}^0 – C_{2019}^1.x + C_{2019}^2.{x^2} + … + C_{2019}^{2018}.{x^{2018}} – C_{2019}^{2019}.{x^{2019}}\end{array}$ Khi đó: $\begin{array}{l}{\left( {1 + x} \right)^{2019}} + {\left( {1 – x} \right)^{2019}} = 2\left( {C_{2019}^0 + C_{2019}^2.{x^2} + … + C_{2019}^{2018}.{x^{2018}}} \right)\\ \Rightarrow C_{2019}^0 + C_{2019}^2.{x^2} + … + C_{2019}^{2018}.{x^{2018}} = \dfrac{{{{\left( {1 + x} \right)}^{2019}} + {{\left( {1 – x} \right)}^{2019}}}}{2} (*)\end{array}$ Thay $x=2$ vào $(*)$ ta có: $\begin{array}{l}C_{2019}^0 + C_{2019}^2{.2^2} + … + C_{2019}^{2018}{.2^{2018}} = \dfrac{{{{\left( {1 + 2} \right)}^{2019}} + {{\left( {1 – 2} \right)}^{2019}}}}{2}\\ \Rightarrow C_{2019}^0 + C_{2019}^2{.2^2} + … + C_{2019}^{2018}{.2^{2018}} = \dfrac{{{3^{2019}} – 1}}{2}\end{array}$Như vậy: $\begin{array}{l}S = 4040\left( {C_{2019}^0 + C_{2019}^2{{.2}^2} + C_{2019}^4{{.2}^4} + … + C_{2019}^{2018}{{.2}^{2018}}} \right)\\ \Rightarrow S = 4040.\dfrac{{{3^{2019}} – 1}}{2}\\ \Rightarrow S = 2020\left( {{3^{2019}} – 1} \right)\end{array}$ Vậy $S = 2020\left( {{3^{2019}} – 1} \right)$ Bình luận
Giải thích các bước giải:
Ta có:
$S = 2.C_{2020}^1 + {3.2^3}.C_{2020}^3 + … + {2019.2^{2019}}.C_{2020}^{2019}$
Số hạng tổng quát của dãy tổng $S$ là:
$\begin{array}{l}
k{.2^k}.C_{2020}^k\left( {1 \le k \le 2019;k\not \vdots 2} \right)\\
= {2^k}.k.C_{2020}^k\\
= {2^k}.2020.C_{2019}^{k – 1}\\
= 4040.C_{2019}^{k – 1}{.2^{k – 1}}
\end{array}$
Khi đó:
$S = 4040\left( {C_{2019}^0 + C_{2019}^2{{.2}^2} + C_{2019}^4{{.2}^4} + … + C_{2019}^{2018}{{.2}^{2018}}} \right)$
Ta xét khai triển:
$\begin{array}{l}
+ ){\left( {1 + x} \right)^{2019}} = C_{2019}^0 + C_{2019}^1.x + C_{2019}^2.{x^2} + … + C_{2019}^{2018}.{x^{2018}} + C_{2019}^{2019}.{x^{2019}}\\
+ ){\left( {1 – x} \right)^{2019}} = C_{2019}^0 – C_{2019}^1.x + C_{2019}^2.{x^2} + … + C_{2019}^{2018}.{x^{2018}} – C_{2019}^{2019}.{x^{2019}}
\end{array}$
Khi đó:
$\begin{array}{l}
{\left( {1 + x} \right)^{2019}} + {\left( {1 – x} \right)^{2019}} = 2\left( {C_{2019}^0 + C_{2019}^2.{x^2} + … + C_{2019}^{2018}.{x^{2018}}} \right)\\
\Rightarrow C_{2019}^0 + C_{2019}^2.{x^2} + … + C_{2019}^{2018}.{x^{2018}} = \dfrac{{{{\left( {1 + x} \right)}^{2019}} + {{\left( {1 – x} \right)}^{2019}}}}{2} (*)
\end{array}$
Thay $x=2$ vào $(*)$ ta có:
$\begin{array}{l}
C_{2019}^0 + C_{2019}^2{.2^2} + … + C_{2019}^{2018}{.2^{2018}} = \dfrac{{{{\left( {1 + 2} \right)}^{2019}} + {{\left( {1 – 2} \right)}^{2019}}}}{2}\\
\Rightarrow C_{2019}^0 + C_{2019}^2{.2^2} + … + C_{2019}^{2018}{.2^{2018}} = \dfrac{{{3^{2019}} – 1}}{2}
\end{array}$
Như vậy:
$\begin{array}{l}
S = 4040\left( {C_{2019}^0 + C_{2019}^2{{.2}^2} + C_{2019}^4{{.2}^4} + … + C_{2019}^{2018}{{.2}^{2018}}} \right)\\
\Rightarrow S = 4040.\dfrac{{{3^{2019}} – 1}}{2}\\
\Rightarrow S = 2020\left( {{3^{2019}} – 1} \right)
\end{array}$
Vậy $S = 2020\left( {{3^{2019}} – 1} \right)$