cho 1phan 2mu2+1phan3mu2+1phan4mu2+....+1phan9mu2 cmr2phan5<s<8phan9
\(S=\dfrac{2mu2}{1.2}+\dfrac{2mu2}{2.3}+\dfrac{2mu2}{3.4}+...+\dfrac{2mu2}{2022.2023}\)
(mu = mũ)
\(S=\dfrac{2^2}{1.2}+\dfrac{2^2}{2.3}+\dfrac{2^2}{3.4}+...+\dfrac{2^2}{2022.2023}\)
\(S=2^2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)\)
\(S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)
\(S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2023}\right)\)
\(S=2^2.\dfrac{2022}{2023}\)
\(S=\dfrac{2^2.2022}{2023}=\dfrac{8088}{2023}\)
Cho b=1+2+2mu2+2mu3+...+2mu6,A=2mu2+2mu3+2mu4+..+2mutam chứng minh rằng A=4B
\(B=1+2+2^2+...+2^6.\)
\(=>4B=2^2+2^3+...+2^8\)\(\left(1\right)\)
\(A=2^2+2^3+...+2^8\)\(\left(2\right)\)
Từ (1) và (2)
=> A = 4B
A=3mu2+3mu2+2mu2+2mu2+.........=1mu2+1mu2
tính:
2mu100-2mu99-....................-2mu2-2mu1
mu=mũ
ai đúng mik tĩk cho
tính:
2mu100-2mu99-....................-2mu2-2mu1
mu=mũ
ai đúng mik tĩk cho
\(A=2^{100}-2^{99}-...-2^2-2\)
\(2A=2^{101}-2^{100}-...-2^3-2^2\)
\(2A-A=2^{101}-2^{100}-...-2^3-2^2-2^{100}+2^{99}+...2^2+2\)
\(A=2^{101}-\left(2^{100}-2^{100}+2^{99}-2^{99}+...+2^2-2^2+-2\right)\)
\(A=2^{101}+2\)
Cho a=1 Phan 11 +1phan 12+1phan 13+...+1phan 70
Chứng minh A>4phan 3
cho s = 1+2+2mu2+...+2mu2018 tính s
22s=2+22+...+22020
4S-S=(2+22+...+22020)-(1+2+22+....+22018)
3S=22020-1
S=(22020-1):3
\(S=1+2+2^2+...+2^{2018}\)
\(\Rightarrow2S=2+2^2+2^3+...+2^{2019}\)
\(\Rightarrow2S-S=\left(2+2^2+2^3+...+2^{2019}\right)-\left(1+2+2^2+...+2^{2018}\right)\)
\(\Rightarrow S=2^{2019}-1\)
Vậy...(tự kết luận)
:)))
1- 1 phan 2 - 1 phan 2 ^2 - 1phan 2^3 - 1phan 2^4-....- 1phan 2^10 .> 1phan 2^11
Biết rằng 1mu 2 +2mu2 +3mu2+......+10mu2= 385 tính nhanh di tổng. S=2mu2+4mu2+6mu2+.....+20mu2
S = 22 + 42 + 62 + ... + 202
= (2.1)2 + (2.2)2 + (2.3)2 ... (2.10)2
= 22.12 + 22.22 + 22.32 + ... + 22.102
= 22 (12 + 22 + ... + 102 )
= 4 . 385
= 1540
1phan 5 công 1phan 5 cộng 1phan 5 bằng bao nhiêu