Is there another way to do this? (1 Viewer)

[ND]

Can the sum 1+11+111+...+11111... (n digits) be found using a method other than this:

1111... (n digits) is the sum of the GP 1+10+100+...(n terms) = (10^n - 1)/9
.'. 1+11+111+...+11111... = (10 - 1)/9 + (10^2 - 1)/9 +.... (10^n - 1)/9
= (10 + 10^2 +...+ 10^n)/9 - n/9
= 10(10^n - 1)/81 - n/9 (as 10 + 10^2 +...+ 10^n is a GP = 10(10^n - 1)/9)

Thanks.

 

[OLDMAN]

Hi ND. Try this.

1+11+111+...+111...1 (n 1's)=
(10/9-1/9)+(100/9-1/9)+(1000/9-1/9)+...+(10^n/9-1/9)=
1/9(10+100+...+10^n)-n/9=
1/9(10(10^n-1)/9)-n/9=(10/81)*(10^n-1)-n/9

 

[OLDMAN]

Dividing by 9 works as well with 2+22+222+..., as well with 3+33+.., or k+kk+kkk+....

 

[ND]

Originally posted by OLDMAN
Dividing by 9 works as well with 2+22+222+..., as well with 3+33+.., or k+kk+kkk+....

Thanks. But for these wouldn't it just be easier to just take the number out the front, and proceed with 1's? What i mean is, k+kk+kkk+... = k(1+11+111+...)

 

[OLDMAN]

You are right.