Question

How we find the sum of geometric sequence ( 1+1+m+m^2 + m^3 + ...+m^n ) ?

Answer 1

please make sure you have no mistakes in it

Answer 2

but, in general,

\(\large\color{black}{ \displaystyle {\rm S}=\frac{a_1}{1-{\rm r}} }\) if the sequence continues forever.

Answer 3

should be just

\(\large\color{black}{ \displaystyle (1+\rm m+m^2+m^3+m^4+~..m^n }\)

for n'th number of terms,

\(\large\color{black}{ \displaystyle {\rm S}_n=\frac{a_1(1-{\rm r}^n)}{1-{\rm r}} }\)

Answer 4

the idea is that when the terms go on forever (and |r|<1, because otherwise it diverges), then r^n approaches 0, and thus (1-r^n) is 1, and 1 times \(a_1\) is just \(a_1\).

Answer 5

however for any n number of terms, again, use:

\(\large\color{black}{ \displaystyle {\rm S}=\frac{a_1(1-{\rm r}^n)}{1-{\rm r}} }\)

Answer 6

should be:

\(\large\color{black}{ \displaystyle {\rm S}_n=\frac{a_1(1-{\rm r}^n)}{1-{\rm r}} }\)

Answer 7

you are multiplying times m every time, so your common ratio is m

Answer 8

Here the original question :#
i got the sequence from substitute

Answer 9

fur ein, in german means for any?

Answer 10

yes

Answer 11

the question is : is it bounded or not and if it converge, what the limit ?..


from my solution , i got that div when | m | >=1 , and con when |m| < 1
but not sure if it true

Answer 12

wait, so it is an INFINITE series, correct?

Answer 13

not just till some a^n, but till a^n as n approaches infinty.
right?

Answer 14

ok, so we will see what this would be like
--------------------------------------
you are given that:
\(\large\color{black}{ \displaystyle {\rm a}_0=1 }\)
\(\large\color{black}{ \displaystyle {\rm a}_{n+1}={\rm a}_n+{\rm m}^n }\)
--------------------------------------
\(\large\color{black}{ \displaystyle {\rm a}_0=1 }\)
\(\large\color{black}{ \displaystyle {\rm a}_1={\rm a}_0+{\rm m}^0=1+{\rm 1}=2 }\)
\(\large\color{black}{ \displaystyle {\rm a}_2={\rm a}_1+{\rm m}^1=2+{\rm m} }\)
\(\large\color{black}{ \displaystyle {\rm a}_3={\rm a}_2+{\rm m}^2=2+{\rm m} +{\rm m^2}}\)
\(\large\color{black}{ \displaystyle {\rm a}_4={\rm a}_3+{\rm m}^3=2+{\rm m}+{\rm m}^2 +{\rm m^3}}\)
so from that point and on, you are adding another \({\rm m}^n\).


\(\large\color{black}{ \displaystyle {\rm Series}=\left( {\rm a}_0\right)+\left( {\rm a}_1\right)+\left( {\rm a}_2\right)+\left( {\rm a}_3\right)+\left( {\rm a}_4\right)+\left( {\rm a}_5\right)+... }\)


\(\large\color{black}{ \displaystyle {\rm Series}=\left( 1\right)+\left( 2\right)+\left( 2+{\rm m}\right)+\left( 2+{\rm m}+{\rm m}^2\right)+ \\[0.5em] \displaystyle \left( 2+{\rm m}+{\rm m}^2+{\rm m}^3\right)+\left( 2+{\rm m}+{\rm m}^2+{\rm m}^3+{\rm m}^4\right)+... }\)


or,


\(\large\color{black}{ \displaystyle {\rm Series}=1+2n+{\rm m}(n-1)+{\rm m^2}(n-2) +{\rm m^3}(n-3)+ \\ \displaystyle {\rm m^4}(n-4)+{\rm m^5}(n-5)+{\rm m^6}(n-6)}\)

Answer 15

i solved it like that , can be true?:D