Sum of series 3 + 33 + 333 + ....��..+ n

If you observe closely , it looks like a Geometric Progression, it is not. First of all ,

we have to convert into a G.P. and then derive the formula.

Step 1: 3[ 1+11+111+.....….n]  ( Take out 3)

Step 2 : (3/9)×9 [ 1+11+111+.....n]  ( multiply and divide by 9 )

Step 3: 3/9 [ 9+99+999+...+n] ( take 9 inside the bracket and multiply with each term)

Step 4: 3/9 [ (10-1)+(100-1)+(1000-1)+.......n]

Step5: 3/9[ 10 + 100 + 1000 + .....n ] - [ 1 + 11 +111 + .…....n ]

Now you can see that series (10+100+ 1000 +......n) is in GP with a= 10 and r=10

Step6: 3/9[ 10 (10^n - 1)/(10-1) - (1+1+1+....n).  {(1+1+1.....n) = 1×n =n }

Step7: 3/9 [ 10 (10^n - 1 )/(9) - n ]

Step8: 3/9 [ 10 ( 10^n - 1) - 9n ) /(9) ] ( by taking 9 as LCM )

Step 9: 3/81 [ 10^n+1 - 10 - 9n ] ( 10 ×10^n = 10^n+1) This is the required formula.

Let us verify the formula by adding say first three terms  (3+33+333 = 369)

Now use the formula Sn = 3/81 [ 10^n+1 - 9n - 10 ) ( with n = 3 ie first three terms )

Sn = 3/81 [ 10^4 - (9 ×3) - 10 ) ]

= 3/81 [ 10000 - 27 - 10 ]

= 3/81 [ 10000 - 37 ] = 369 . Hence verified.

You can use this method to sum up similar sequences like , 0.5 + 0.55 + 0.555 +....