chứng minh rằng 1/2019C1 +1/2019C2+..+1/2019C2019=1010/2019(1/2018C0+1/2018C1+..+1/2018C2018) giúp gấp ạ 15/08/2021 Bởi Alaia chứng minh rằng 1/2019C1 +1/2019C2+..+1/2019C2019=1010/2019(1/2018C0+1/2018C1+..+1/2018C2018) giúp gấp ạ
Giải thích các bước giải: Ta có :$VT=\dfrac{1}{2019}(\dfrac{1}{C^0_{2018}}+\dfrac{2}{C^1_{2018}}+..+\dfrac{2019}{C^0_{2018}})$ Mà $C^k_n=C^{n-k}_n$ $\to VT=\dfrac{1}{2019}(\dfrac{1+2019}{C^0_{2018}}+\dfrac{2+2018}{C^1_{2018}}+..+\dfrac{1008+2012}{C^{1007}_{2018}}+\dfrac{1010}{C^{1008}_{2018}})$ $\to VT=\dfrac{1}{2019}(\dfrac{2020}{C^0_{2018}}+\dfrac{2020}{C^1_{2018}}+..+\dfrac{2020}{C^{1007}_{2018}}+\dfrac{1010}{C^{1008}_{2018}})$ $\to VT=\dfrac{1010}{2019}(\dfrac{2}{C^0_{2018}}+\dfrac{2}{C^1_{2018}}+..+\dfrac{2}{C^{1007}_{2018}}+\dfrac{1}{C^{1008}_{2018}})$ $\to VT=\dfrac{1010}{2019}(\dfrac{1}{C^0_{2018}}+\dfrac{1}{C^{2018}_{2018}}+\dfrac{1}{C^1_{2018}}+\dfrac{1}{C^{2017}_{2018}}+..+\dfrac{1}{C^{1008}_{2018}})$ $\to VT=\dfrac{1010}{2019}(\dfrac{1}{C^0_{2018}}+\dfrac{1}{C^1_{2018}}+..+\dfrac{1}{C^{2018}_{2018}})$ $\to dpcm$ Bình luận
Giải thích các bước giải:
Ta có :
$VT=\dfrac{1}{2019}(\dfrac{1}{C^0_{2018}}+\dfrac{2}{C^1_{2018}}+..+\dfrac{2019}{C^0_{2018}})$
Mà $C^k_n=C^{n-k}_n$
$\to VT=\dfrac{1}{2019}(\dfrac{1+2019}{C^0_{2018}}+\dfrac{2+2018}{C^1_{2018}}+..+\dfrac{1008+2012}{C^{1007}_{2018}}+\dfrac{1010}{C^{1008}_{2018}})$
$\to VT=\dfrac{1}{2019}(\dfrac{2020}{C^0_{2018}}+\dfrac{2020}{C^1_{2018}}+..+\dfrac{2020}{C^{1007}_{2018}}+\dfrac{1010}{C^{1008}_{2018}})$
$\to VT=\dfrac{1010}{2019}(\dfrac{2}{C^0_{2018}}+\dfrac{2}{C^1_{2018}}+..+\dfrac{2}{C^{1007}_{2018}}+\dfrac{1}{C^{1008}_{2018}})$
$\to VT=\dfrac{1010}{2019}(\dfrac{1}{C^0_{2018}}+\dfrac{1}{C^{2018}_{2018}}+\dfrac{1}{C^1_{2018}}+\dfrac{1}{C^{2017}_{2018}}+..+\dfrac{1}{C^{1008}_{2018}})$
$\to VT=\dfrac{1010}{2019}(\dfrac{1}{C^0_{2018}}+\dfrac{1}{C^1_{2018}}+..+\dfrac{1}{C^{2018}_{2018}})$
$\to dpcm$