Question

prove that the sum of n terms of the sequence (9,99,999,.............) equals 10/9(10 to the power of n -1 )-n

Answer 1

Are you trying to show it equals \[\large {10 \over 9} \cdot (10^{n-1})^{-n}\]

Answer 2

yes but the -1 and -n are down not powers

Answer 3

so \[{10 \over 9} \cdot (10^n -1)-n\]

Answer 4

yes thats it exactly

Answer 5

so what do u think ?

Answer 6

Use 9 as your first number, then you add\[9*1+ 9*11+9*111+9*1111+...\]

Answer 7

thanks

Answer 8

Then factor out a 9, so you have \[9(1+11+111+1111+...)\]

Answer 9

And this can be rewritten as \[9[(10^0 +1)+(10^1+10^0)+(10^2+10^1+10^0)+...+(10^{n-1}+...+10^1+10^0)]\]

Answer 10

Using some more manipulation of sums of geometric series, you should be able to reach your answer.

Answer 11

thank u :D

Answer 12

You're welcome.

Answer 13

u r smart :D

Answer 14

I've just been around the math block (so to speak) so I've seen a number of these tricks. But thanks for the compliment! :)

Answer 15

:D welcome