Question

What is the sum of the series to the nth term? \[2.4+2.44+2.444+2.4444+......\]

Answer 1

Is there any one who try this before I post the answer?

Answer 2

@Jonask what did you say?

Answer 3

there is a way we write these in fraction form\[2,444444....=x\]

Answer 4

i dont knw if that can help

Answer 5

let \[x=2,4\]

Answer 6

forget 2
0.4=4/10,0.44=44/100, 0.444=444/1000
correct way ?

Answer 7

yape @hartnn

Answer 8

Then what? @hartnn

Answer 9

11/10 is common ratio, isn't it ?

Answer 10

nope

Answer 11

oh, yeah...it isn't.....then i m stuck

Answer 12

will I post the answer then?

Answer 13

ok.

Answer 14

\[2.4+2.44+2.444+2.4444+......\]
\[2+0.4+2+0.44+2+0.444+........\]\[2n+0.4+0.44+0.444+........\]\[2n+\frac{ 4 }{ 9 }(1-\frac{ 1 }{ 10 })+\frac{ 4 }{ 9 }(1-\frac{ 1 }{ 10^{2} })+\frac{ 4 }{ 9 }(1-\frac{ 1 }{ 10^{3} })+........+\frac{ 4 }{ 9 }(1-\frac{ 1 }{ 10^{n} })\]\[2n+\frac{ 4 }{ 9 }(1-\frac{ 1 }{ 10 }+1-\frac{ 1 }{ 10^{2} }+1-\frac{ 1 }{ 10 ^{3}}+.........+1-\frac{ 1 }{ 10 ^{n}})\]\[2n+\frac{ 4 }{ 9 }(n-(\frac{ 1 }{ 10 }+\frac{ 1 }{ 10^{2} }+\frac{ 1 }{ 10 ^{3}}+.........+\frac{ 1 }{ 10 ^{n}}))\]\[2n+\frac{ 4 }{ 9 }(n-(\frac{ \frac{ 1 }{ 10 }(1-\frac{ 1 }{ 10^{n} }) }{ 1-\frac{ 1 }{ 10 } }))\] Thus finally \[\frac{ 22n }{ 9 }-\frac{ 4 }{ 81 }(1-\frac{ 1 }{ 10^{n} })\] will be the answer.

Answer 15

ohh!! (1-1/10)

Answer 16

\[(x)+(x+0,4)+(x+0,4+0,04)+(x+0,4+0,04+0,004)+....\]
\[=x(n+1)+ 0,4(n)+\sum_{k=1}^{n}\ (0,4)(0,1)^{n} \]

Answer 17

x=2