Question

Find the sum of the series 5+55+555+... to n terms, let f(n) be Sn, then verify that S2=60 when n=2 using f(n), and lastly, establish that 81/x f(n) is always a multiple of 81

Answer 1

\[5*\sum_{1}^{n}n10^{^n-1}\]

This is what I got, do you think it is right?

Answer 2

that is n*10^(n-1)

Answer 3

LOLs, sorry i dnt know how to that, havent learnt it yet xD

Answer 4

I found the rule that
1st term 5*1
2nd 5*12
3rd 5*123
4th 5*1234

Answer 5

substitute for n = 2 in the formula
= 5 * (2 * 10^1) = 100

Answer 6

oh yeah it is wrong

Answer 7

afraid so - i'm taking a look - but this ones a bit tricky

Answer 8

and it might be that the rule will change after 123456789

Answer 9

To make the formula easier to see you can factor a 5 out:\[5(1+11+111+1111+\ldots)\]

Answer 10

Then the number 111...111, where there are n ones is:
\[111\ldots 11 = \sum_{i=0}^{n-1}10^i\]

Thats a geometric sum. The answer is:
\[\sum_{i=0}^{n-1}10^i = \frac{10^n-1}{10-1}\]

Answer 11

that is the last number

Answer 12

hey joe! this is not a geometric sum!

Answer 13

what isnt a geometric sum? The 111...111?

Answer 14

yes

Answer 15

yes it is. its:
\[1+10+100+1000+\ldots\]

Answer 16

oh no u are right

Answer 17

r = 10, first term is 1.

Answer 18

yep, sorry

Answer 19

So im a little confused das to what is the question we are trying to answer.

Answer 20

:-)

Answer 21

We are trying to show that this sum is always divisible by 81?

Answer 22

find a formula for 5+55+555....

Answer 23

oh ok. Well the formula for the nth term is:
\[a_n = 5\frac{10^n-1}{10-1} = \frac{5}{9}(10^n-1)\]So we are going to sum those terms:
\[\sum_{i=0}^{n-1}\frac{5}{9}(10^i-1) = \frac{5}{9}\left(\sum_{i=0}^{n-1}10^i-\sum_{i=0}^{n-1}1\right)\]\[=\frac{5}{9}\left(\frac{10^n-1}{10-1}-n\right) = \frac{5}{81}(10^n-9n-1)\]

Answer 24

That is what I started doing as well but could not finish
U are ace!

Answer 25

i thinki might have messed up on my index somewhere...

Answer 26

because if i plug in n = 2, i get the sum of only the first term, not the first 2 terms...arg.

Answer 27

Just change the n's to n+1's to fix that.

Answer 28

Yeah, the correct formula should be:\[\frac{5}{81}(10^{n+1}-9(n+1)-1) = \frac{5}{81}(10^{n+1}-9n-10)\]sry bout that.

Answer 29

that looks good