cho biểu thức A=1+5+52+53+ ............................+52017
tìm số tự nhiên n biết 4a+1=5n+1
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
30 tháng 7 2023
A= 1 + 5 + 52 + 5 3 + ... + 5800
5A= 5 + 52 + 53 + .... +5 800 + 5801
5A - A = 5801 - 1
4a = 5801 - 1
5801 - 1 +1 = 5n
⇒ 5801 = 5n ⇒ n = 801
LH
26 tháng 10 2017
A = 1 + 5 + 52 + 53 + ....+ 52017
A . 5 = 5 + 52 + 53 + 54 + .... + 52018
A . 5 - A = ( 5 + 52 + 53 + 54 + .... + 52018 ) - ( 1 + 5 + 52 + 53 + ......+ 52017 )
A . 4 = 52018 - 1
Ta có : 52018 - 1 + 1 = 5n + 1
52018 = 5n+1
Suy ra : 2018 = n + 1
2018 - 1 = n
2017 = n
26 tháng 10 2017
chuẩn mình cũng làm thế
đó là đề thi khảo sát giữa học kì 1
NH
0
26 tháng 3 2021
Bài 2:
a) Ta có: \(A=\dfrac{4}{n-1}+\dfrac{6}{n-1}-\dfrac{3}{n-1}\)
\(=\dfrac{4+6-3}{n-1}\)
\(=\dfrac{7}{n-1}\)
Để A là số tự nhiên thì \(7⋮n-1\)
\(\Leftrightarrow n-1\inƯ\left(7\right)\)
\(\Leftrightarrow n-1\in\left\{1;7\right\}\)
hay \(n\in\left\{2;8\right\}\)
Vậy: \(n\in\left\{2;8\right\}\)
27 tháng 3 2021
ta có B=2n+9/n+2-3n+5n+1/n+2=4n+10/n+2 Để B là STN thì 4n+10⋮n+2 4n+8+2⋮n+2 4n+8⋮n+2 ⇒2⋮n+2 n+2∈Ư(2) Ư(2)={1;2} Vậy n=0
NH
0
DT
1
4 tháng 3 2021
a) \(4n-5⋮13\)
\(\Rightarrow4n-5+13⋮13\Rightarrow4n+8⋮13\Rightarrow4\left(n+2\right)⋮13\)
Vì (4;13) = 1 nên n+2 chia hết cho 13
=> n=13k-2 ( \(k\in N\)*)
b) \(5n+1⋮7\Rightarrow5n+1+14⋮7\Rightarrow5n+15⋮7\Rightarrow5\left(n+3\right)⋮7\)
Vì 5 không chia hết cho 7 nên để 5(n+3) chia hết cho 7 thì n+3 chia hết cho 7
=> n = 7k-3 ( \(k\in N\)*)
c) \(25n+3⋮53\Rightarrow25n+3-53⋮53\Rightarrow25n-50⋮53\Rightarrow25\left(n-2\right)⋮53\Rightarrow n-2⋮53\)
=> n = 53k+2 ( k thuộc N*)
29 tháng 1 2017
a,
4n - 5 \(⋮\)13
=> 4n - 5 + 13 \(⋮\)13
=> 4n + 8 \(⋮\)13
=> 4.(n+2)\(⋮\)13
=> n + 2 \(⋮\)13
=> n +2 = 13k ( k\(\in\)N*)
=> n = 13k - 2
vậy: n = 13k - 2 ( k\(\in\)N*)
b, 5n + 1 \(⋮\)7
=> 5n + 1 + 14 \(⋮\)7
=> 5n + 15 \(⋮\)7
=> 5. ( n+3) \(⋮\)7
=> n + 3 \(⋮\)7
=> n+3 = 7k ( k\(\in\)N*)
=> n = 7k - 3
vậy: n = 7k - 3 ( k\(\in\)N*)
c, 25n + 3 \(⋮\)53
phần c thì mk chịu. bạn tk mk nha. 2 phần kia đúng 100%
TT
4
16 tháng 7 2016
\(A=5^{2016}-5\)
A=132901150760150400933474662701093632444139156230245797983451739952061292318821219082408733380123716446923280138816148691348332250549138432694744733040207471635460062291111714453852983450163412839478432674285466723489853471331344961752024356711039744998722729088056022242066820496791634992123859739046602165056020296822649485842368328334914700117232737216924944154499322138498785527017914889599336202481672782191035035874706832781528727280801652013578429369125463744179027114136759472454584397133928400078670849997607302892223027036473470262496682733564340461161715868386687990733274505753924648948618963125139438987342574828670180634045054186337242659614976824201571903960747675319866959366451316077662320815346287052220792434027927921187356889656584951394657674940726699259495071241216158196484638282891605536718919121672173792307092698308883330916383232492806602360867087932017350554747339691684066271395957046064307027329280820284160155505133882385577240294382888635735834661135764449778633852155557799373087364612366519453980045038199609836307800276035054500661361991243746011829792746699524810528841093775444529181087096473054405737871791062821700667456513545082416389778381211311121521088261300886212326120546085043586116353533714697985212811857529689920199233762425541566473083922473532034610100101045817053433299648552633995654263623546743263019492984489331442211901279648600393556989729404449462890620
b, n = 106320920608120320746779730160874905955311324984196638386761391961649033855056975265926986704098973157538624111052918953078665800439310746155795786432165977308368049832889371563082386760130730271582746139428373378791882777065075969401619485368831795998978183270444817793653456397433307993699087791237281732044816237458119588673894662667931760093786189773539955323599457710799028421614331911679468961985338225752828028699765466225222981824641321610862743495300370995343221691309407577963667517707142720062936679998085842313778421629178776209997346186851472368929372694709350392586619604603139719158895170500111551189874059862936144507236043349069794127691981459361257523168598140255893567493161052862129856652277029641776633947222342336949885511725267961115726139952581359407596056992972926557187710626313284429375135297337739033845674158647106664733106585994245281888693670345613880443797871753347253017116765636851445621863424656227328124404107105908461792235506310908588667728908611559822907081724446239498469891689893215563184036030559687869046240220828043600529089592994996809463834197359619848423072875020355623344869677178443524590297432850257360533965210836065933111822704969048897216870609040708969860896436868034868893082826971758388170249486023751936159387009940433253178467137978825627688080080836653642746639718842107196523410898837394610415594387591465153769521023718880314845591783523559570312500
\(A=1+5+5^2+...+5^{2017}\)
\(5A=5.\left(1+5+...+5^{2017}\right)\)
\(5A=5+5^2+5^3+...+5^{2018}\)
\(5A-A=5+5^2+...+5^{2018}-1-5-5^2-...-5^{2017}\)
\(4A=5^{2018}-1\)
Thay \(4A=5^{2018}-1\)vào 4 + 1 = 5n+1, ta có:
\(5^{2018}-1+1=5^{n+1}\)
\(\Rightarrow5^{2018}=5^{n+1}\Rightarrow n+1=2018\Rightarrow n=2017\)