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\(\frac{1}{5}A=\frac{1}{5}.\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{20}}\right)\)
\(\Rightarrow\frac{1}{5}A=\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{20}}\)
\(\Rightarrow\frac{1}{5}A-A=\left(\frac{1}{5^2}+...+\frac{1}{5^{21}}\right)-\left(\frac{1}{5}+...+\frac{1}{5^{20}}\right)\)
\(-\frac{4}{5}A=\frac{1}{5^{21}}-\frac{1}{5}\)
\(\Rightarrow A=\left(\frac{1}{5^{21}}-\frac{1}{5}\right):\left(-\frac{4}{5}\right)\)
các câu còn lại tương tự thôi
B1 c2
dùng xích ma \(\text{∑}^{20}_1\left(\frac{1}{5^x}\right)=0,25=\frac{1}{4}\)
chỗ phía dưới là 1 nha nó bị che
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+....+\frac{19}{9^2.10^2}=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{9^2}-\frac{1}{10^2}=1-\frac{1}{10^2}
\(A=\dfrac{3}{1^2+2^2}+\dfrac{5}{2^2+3^2}+...+\dfrac{19}{9^2+10^2}\) (sửa \(1^22^2\) thành \(1^2+2^2\))
Ta có : \(\left(1+2\right)^2=1^2+2^2+2.1.2\Rightarrow1^2+2^2< \left(1+2\right)^2\)
\(\Rightarrow1^2+2^2< 3^2=3.3\)
\(\Rightarrow\dfrac{3}{1^2+2^2}< \dfrac{1}{3}< 1\)
Tương tự \(\dfrac{5}{2^2+3^2}< \dfrac{1}{5}< 1\)
\(.....\)
\(\dfrac{9}{9^2+10^2}< \dfrac{1}{19}< 1\)
\(\Rightarrow A=\dfrac{3}{1^2+2^2}+\dfrac{5}{2^2+3^2}+...+\dfrac{19}{9^2+10^2}< 1.9=9< 1\)
\(\Rightarrow dpcm\)
Bài 1:
Ta có:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Mà \(\frac{99}{100}< 1\)
\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)
\(B=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+....+\frac{19}{9^2.10^2}\)
\(B=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+....+\frac{19}{81.100}\)
\(B=\frac{4-1}{1.4}+\frac{9-4}{4.9}+\frac{16-9}{9.16}+....+\frac{100-81}{81.100}\)
\(B=\frac{4}{1.4}-\frac{1}{1.4}+\frac{9}{4.9}-\frac{4}{4.9}+\frac{16}{9.16}-\frac{9}{9.16}+...+\frac{100}{81.100}-\frac{81}{81.100}\)
\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+....+\frac{1}{81}-\frac{1}{100}\)
\(B=1-\frac{1}{100}< 1\)
=> B < 1 (Đpcm)
B = 3/12.22 + 5/22.32 + 7/32.42 + ... + 19/92.102
B = 3/1.4 + 5.4.9 + 7/9.16 + ... + 19/81.100
B = 1 - 1/4 + 1/4 - 1/9 + 1/9 - 1/16 + ... + 1/81 - 1/100
B = 1 - 1/100 < 1 ( đpcm)
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{19}{81.100}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{81}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}< 1\left(dpcm\right)\)
câu 1:
2 + 2^2 + 2^3 + ... + 2^20 = 2( 1 + 2 + 2^2 +... + 2^19) chia hết cho 2
câu 2
2 + 2^2 + 2^3 + 2^4 +... + 2^19 + 2^20
= ( 2 + 2^2) + ( 2^3 + 2^4) + ....+ ( 2^19 + 2^20)
= 2( 1 + 2 ) + 2^3( 1+3) +...+ 2^19(1+2)
= 2. 3 + 2^3 . 3 +...+2^19.3
= 3.(2+2^3+2^5+....+2^19) chia hết cho 3
\(a.2+2^2+2^3+...+2^{19}\)\(+2^{20}\)
Ta có: \(2⋮2,2^2,2^3⋮2,..2^{19}⋮2,2^{20}⋮2\)
\(\Rightarrow2+2^2+2^3+...+2^{19}+2^{20}⋮2\)
b.Giống trên