Question

help me out plz...

Answer 1

\[1+ 1/5+4/5^{2}+9/5^{3}+......\]

Answer 2

\[\frac{ n^2 }{ 5^n }\]
?

what's your question?

Answer 3

summation
till infinity

Answer 4

Do you know summation notation?

Answer 5

\[1+\sum_{1}^{\inf} n^{2}/5^{n}\]

Answer 6

@Algebraic!

I'm terrible with Summation, can you take it from here?

Answer 7

the n^2 part is easy
the other part is a geometric series
use the sum of products is product of sums, I believe ...

Answer 8

i dint get you???

Answer 9

\[\sum_{i=0}^{n} i ^{2} =\frac{ n(n+1)(2n+1) }{6 }\]

Answer 10

i know that, but how are you going to apply it here??

Answer 11

\[\sum_{i=0}^{n} r^{i} =\frac{1 }{1-r }\]

Answer 12

are you suggesting that we should multiply these two ??

Answer 13

you can't multiply series like that!

Answer 14

not sure... that doesn't seem to give much of an answer...

Answer 15

need two different indices...?

Answer 16

such a weird summation, and i used a calculator and gave up once i reached n=21

Answer 17

anyways i got the solution. thank you guys...

Answer 18

you just needed an approximation??

Answer 19

nope..
exact answer. it turned out to be 75/32

Answer 20

share your insights...

Answer 21

don't think that's right anyway.

Answer 22

first 6 terms are 1.4625

Answer 23

I think my answer is wrong. but this procedure, I think is good..
say x=1/5.
and the summation is
\[S=x^{1}+4x^{2}+9x^{3}+....\]
lets add the 1 later
\[Sx=x^{2}+4x^{3}+9x^{4}+..\]
\[S(1-x)=x^{1}+3x^{2}+5x^{3}+..\]
\[Sx(1-x)=x^{2}+3x^{3}+5x^{4}+..\]
\[S(1-x)-Sx(1-x)=x^{1}+2x^{2}+2x^{3}+..\]
\[S(1-x)^2=x^{1}+2x^{2}[1+x+x^{2}..]\]
i think you can take it from here......
all we have to do is to find S+1...

Answer 24

should be 1 + 2/(ln5)^3

Answer 25

1.46875, im going with this answer
screwed up my program the first time...

Answer 26

not sure how exactly...

Answer 27

get one of the math guys in here lol...

Answer 28

never fear
@hartnn is here

Answer 29

the expression n^2/5^n converges meaning the fraction becomes smaller and smaller

i really doubt that the summation will ever be larger than .5, mainly cause i wrote a program and ran it till n=144

and it seems to stop at .46875 or in fraction form 15/32

now the question is how do you go about getting that fraction without writing a program to calculate, that i have no idea

Answer 30

"the fraction getting smaller and smaller" is not the same thing as convergence. sum 1/n and sum 1/n^2 both have terms going to 0, but first sum goes to infinity and second to a Pi^2/6,

Answer 31

I think this might work. F(x) = Sum(n^2 * x^n).
F/x = 1/x + 1 + 4x + 9x^2 ...
Int(F/x) = c+ ln(x) + x + 2x^2 + 3x^3 ...
Int(F/x)/x = c/x + ln(x)/x + 1 + 2x + 3x^2
Int( Int(F/x)/x ) = d + c ln(x) + int (ln /x) + x + x^2 + x^3 ...
= d + c ln(x) + int (ln /x) + (1/(1-x) - 1) geometric series
Then you reverse those steps and take derivatives and multiply through by x. and you get and expression for the sum then plug in x = 1/5