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\(S=3^0+3^2+3^4+3^6+...+3^{2002}\)
\(\Rightarrow3^2S=3^2+3^4+3^6+3^8+...+3^{2002}+3^{2004}\)
\(\Rightarrow9S-S=\left(3^2+3^4+3^6+3^8+...+3^{2002}+3^{2004}\right)-\left(3^0+3^2+3^4+3^6+...3^{2000}+3^{2002}\right)\)
\(\Rightarrow8S=3^{2004}-3^0=3^{2004}-1\)
\(\Rightarrow S=\frac{3^{2004}-1}{8}\)
a) S = 30 + 32 + 34 + ..... + 32002
9S = 32 + 34 + ..... + 32002 + 32004
9S - S = (32 + 34 + ..... + 32002 + 32004) - (30 + 32 + 34 + ..... + 32002)
8S = 32004 - 30
S = \(\frac{3^{2004}-1}{8}\)
b) S = 30 + 32 + 34 + ..... + 32002
S = (30 + 32 + 34) + (36 + 38 + 310) + ..... + (32000 + 32001 + 32002)
S = (1 + 9 + 81) + 36.(1 + 9 + 81) + ..... + 32000.(1 + 9 + 81)
S = 91 + 36 . 91 + ...... + 32000 . 91
S = 91 . (1 + 36 + ...... + 32000)
S = 7 . 13 . (1 + 36 + ...... + 32000)
Đặt A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
A=\(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}\)
Ta thấy :
\(\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4.4}< \dfrac{1}{3.4};...;\)
\(\dfrac{1}{100.100}< \dfrac{1}{99.100}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
Nhân xét :
\(\dfrac{1}{1.2}=1-\dfrac{1}{2};\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};\)
\(...;\dfrac{1}{99.100}=\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{4}+...+\)
\(\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow A< 1-\dfrac{1}{100}\)
\(\Rightarrow A< \dfrac{99}{100}\)
Vì \(A< \dfrac{99}{100}< 1\)
\(\Rightarrow A< 1\)
Bài 1)
Đặt \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{100^2}\)
Ta thấy:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4};....;\dfrac{1}{100^2}=\dfrac{1}{100.100}< \dfrac{1}{99.100}\)\(\Rightarrow\) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{100^2}\) < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{99.100}\)
\(\Rightarrow\) A < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow\) A < \(1-\dfrac{1}{100}\) < 1 \(\Rightarrow\) A < 1
Vậy \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{100^2}\)< 1
A) Nhân S với 32 ta được :
9S = 3^2 + 3^4+...+ 3^2002 + 3^2004
\(\Rightarrow\)9S - S = ( 3^2 + 3^4 + .. + 3^2004 ) - ( 3^0 + 3^4+...2^2002 )
\(\Rightarrow\)8S = 32004 - 1
\(\Rightarrow\)S = 32004 - 1 /8
B) Ta có S là số nguyên nên phải chứng minh 32004 - 1 chia hết cho 7
Ta có : 32004 - 1 (36)334 - 1 = ( 36 - 1 ).M =7.104.M
\(\Rightarrow\)32004 chia hết cho 7 . Mặt khác ƯCLN (7;8)= 1 nên S chia hết cho 7
Kết bạn với mình nhé
Cảm ơn bạn nhiều
a, \(S=3^0+3^2+3^4+...+3^{2002}\)
\(3^2S=3^2+3^4+3^6+...+3^{2004}\)
\(8S=3^{2004}-1\)
\(S=\frac{3^{2004}-1}{8}\)
S=30+32+34+36+...+32002
S=(30+32+34)+(36+38+310)+...(31988+32000+32002)
S=91.1+36.(30+32+34)+...+31988.(30+32+34)
S=91.1+36.91+...+31988.91
S=91.(1+36+..+31988)
S=7.13.(1+36+..+31988)
=>S chia hết cho 7
a, \(S=3^0+3^2+3^4+....+3^{2002}\)
\(3S=3+3^3+....+3^{2003}\)
\(2S=3^{2003}-1\)
b, \(S=\left(3^0+3^2+3^4\right)+\left(3^4+3^6+3^8\right)+...+\left(3^{2000}+3^{1998}+3^{2002}\right)⋮7\)
=> (đpcm)
Cho : S = 30 + 32 + 34 + 36 +.......+ 32002
a) Tính S
b) Chứng minh S\(⋮\) 7.
Giúp mk vs Nguyễn Anh Duy
a) \(S=3^0+3^2+3^4+3^6+...+3^{2002}\)
\(\Rightarrow S=1+3^2+3^4+...+3^{2002}\)
\(\Rightarrow9S=3^2+3^4+3^6+...+3^{2004}\)
\(\Rightarrow9S-S=\left(3^2+3^4+3^6+...+3^{2004}\right)-\left(1+3^2+3^4+...+3^{2002}\right)\)
\(\Rightarrow8S=3^{2004}-1\)
\(\Rightarrow S=\frac{3^{2004}-1}{8}\)
b) \(S=3^0+3^2+3^4+3^6+...+3^{2002}\)
\(\Rightarrow S=\left(3^0+3^2+3^4\right)+\left(3^6+3^8+3^{10}\right)+...+\left(3^{2000}+3^{2001}+3^{2002}\right)\)
\(\Rightarrow S=\left(1+9+81\right)+3^6.\left(1+3^2+3^4\right)+...+3^{2000}.\left(1+3^2+3^4\right)\)
\(\Rightarrow S=91+3^6.91+...+3^{2000}.91\)
\(\Rightarrow S=\left(1+3^6+...+3^{2000}\right).91⋮7\)
\(\Rightarrow S⋮7\)
b) Câu này mình có cách khác:
Ta có S là số nguyên nên phải chứng minh \(3^{2004}-1\) chia hết cho 7
Ta có: \(3^{2004}-1=\left(3^6\right)^{334}-1=\left(3^6-1\right).M=728.M=7.104.M\)
\(\Rightarrow3^{2004}\) chia hết cho 7. Mặt khác \(\left(7;8\right)=1\) nên S chia hết cho 7
câu 1hinhf như sai đề
Tớ nghĩ là S= 30 + 32 + 34 +36 +...+ 32002
thì đúng hơn
xin lỗi chị em mới học lớp 5
b; S=(3^0+3^2+3^4)+......+(3^1998+3^200+3^202)
=91+.....3^1998*(1+3^2+3^4)
=91+.....+3^1998*91
=91+.....+3^1998*13*7 => S chia het cho 7