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2802
1773
`1,A=1/5+1/5^2+1/5^3+...+1/5^2024`
`5A=5.(1/5+1/5^2+1/5^3+...+1/5^2024)`
`5A=1+1/5+1/5^2+...+1/5^2023`
`5A-A=(1+1/5+1/5^2+...+1/5^2023)-(1/5+1/5^2+1/5^3+...+1/5^2024)`
`4A=1-1/5^2024`
`A=(1-1/5^2024)/4`
Vậy `A=(1-1/5^2024)/4`
`2,B=1/5-1/5^2+1/5^3-1/5^4+...+1/5^2023-1/5^2024`
`5B=5.(1/5-1/5^2+1/5^3-1/5^4+...+1/5^2023-1/5^2024)`
`5B=1-1/5+1/5^2-1/5^3+...+1/5^2022-1/5^2023`
`5B+B=(1-1/5+1/5^2-1/5^3+...+1/5^2022-1/5^2023)+(1/5-1/5^2+1/5^3-1/5^4+...+1/5^2023-1/5^2024)`
`6B=1-1/5^2024`
`B=(1-1/5^2024)/6`
Vậy `B=(1-1/5^2024)/6`
`3,C=1/3+1/3^3+1/3^5+...+1/3^2023`
`3^2C=3^2.(1/3+1/3^3+1/3^5+...+1/3^2023)`
`9C=3+1/3+1/3^3+...+1/3^2021`
`9C-C=(3+1/3+1/3^3+...+1/3^2021)-(1/3+1/3^3+1/3^5+...+1/3^2023)`
`8C=3-1/3^2023`
`C=(3-1/3^2023)/8`
Vậy `C=(3-1/3^2023)/8`
`4,D=1/3-1/3^4+1/3^7-1/3^10+...+1/3^2021-1/3^2024`
`3^3D=3^3.(1/3-1/3^4+1/3^7-1/3^10+...+1/3^2021-1/3^2024)`
`27D=9-1/3+1/3^4-1/3^7+...+1/3^2018-1/3^2021`
`27D+D=(9-1/3+1/3^4-1/3^7+...+1/3^2018-1/3^2021)+(1/3-1/3^4+1/3^7-1/3^10+...+1/3^2021-1/3^2024)`
`28D=9-1/3^2024`
`D=(9-1/3^2024)/28`
Vậy `D=(9-1/3^2024)/28`
Hãy giúp mọi người biết câu trả lời này thế nào?
2723
1954
Đáp án:
Giải thích các bước giải:
` 1) `
` A = 1/5 + ( 1 )/( 5^2 ) + ( 1 )/( 5^3 ) + . . . + ( 1 )/( 5^2024 ) `
` 5A = 1 + 1/5 + ( 1 )/( 5^2 ) + . . . + ( 1 )/( 5^2023 ) `
` 5A - A = ( 1 + 1/5 + ( 1 )/( 5^2 ) + . . . + ( 1 )/( 5^2023 ) ) - ( 1/5 + ( 1 )/( 5^2 ) + ( 1 )/( 5^3 ) + . . . + ( 1 )/( 5^2024 ) ) `
` 4A = 1 - ( 1 )/( 5^2024 ) `
` 4A = ( 5^2024 - 1 )/( 5^2024 ) `
` A = ( ( 5^2024 - 1 )/( 5^2024 ) )/( 4 ) `
` 2) `
` B = 1/5 - ( 1 )/( 5^2 ) + ( 1 )/( 5^3 ) - ( 1 )/( 5^4 ) + . . . + ( 1 )/( 5^2023 ) - ( 1 )/( 5^2024 ) `
` 5B = 1 - 1/5 + ( 1 )/( 5^2 ) - ( 1 )/( 5^3 ) + . . . + ( 1 )/( 5^2022 ) - ( 1 )/( 5^2023 ) `
` 5B + B = ( 1 - 1/5 + ( 1 )/( 5^2 ) - ( 1 )/( 5^3 ) + . . . + ( 1 )/( 5^2022 ) - ( 1 )/( 5^2023 ) ) + ( 1/5 - ( 1 )/( 5^2 ) + ( 1 )/( 5^3 ) - ( 1 )/( 5^4 ) + . . . + ( 1 )/( 5^2023 ) - ( 1 )/( 5^2024 ) ) `
` 6B = 1 - ( 1 )/( 5^2024 ) `
` 6B = ( 5^2024 - 1 )/( 5^2024 ) `
` B = ( ( 5^2024 - 1 )/( 5^2024 ) )/( 6 ) `
` 3) `
` C = 1/3 + ( 1 )/( 3^3 ) + ( 1 )/( 3^5 ) + . . . + ( 1 )/( 3^2023 ) `
` 3^2 C = 3 + 1/3 + ( 1 )/( 3^3 ) + . . . + ( 1 )/( 3^2021 ) `
` 9C - C = ( 3 + 1/3 + ( 1 )/( 3^3 ) + . . . + ( 1 )/( 3^2021 ) ) - ( 1/3 + ( 1 )/( 3^3 ) + ( 1 )/( 3^5 ) + . . . + ( 1 )/( 3^2023 ) ) `
` 8C = 3 - ( 1 )/( 3^2023 ) `
` 8C = ( 3^2024 - 1 )/( 3^2023 ) `
` C = ( ( 3^2024 - 1 )/( 3^2023 ) )/( 8 ) `
` 4) `
` D = 1/3 - ( 1 )/( 3^4 ) + ( 1 )/( 3^7 ) - ( 1 )/( 3^10 ) + . . . + ( 1 )/( 3^2021 ) - ( 1 )/( 3^2024 ) `
` 3^3 D = 9 - 1/3 + ( 1 )/( 3^4 ) - ( 1 )/( 3^7 ) + . . . + ( 1 )/( 3^2018 ) - ( 1 )/( 3^2021 ) `
` 27D + D = ( 9 - 1/3 + ( 1 )/( 3^4 ) - ( 1 )/( 3^7 ) + . . . + ( 1 )/( 3^2018 ) - ( 1 )/( 3^2021 ) ) + ( 1/3 - ( 1 )/( 3^4 ) + ( 1 )/( 3^7 ) - ( 1 )/( 3^10 ) + . . . + ( 1 )/( 3^2021 ) - ( 1 )/( 3^2024 ) ) `
` 28D = 9 - ( 1 )/( 2^2024 ) `
` 28D = ( 3^2026 - 1 )/( 2^2024 ) `
` D = ( ( 3^2026 - 1 )/( 3^2024 ) )/( 28 ) `
Hãy giúp mọi người biết câu trả lời này thế nào?
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