Addition and summation are simple ~!

First of all, if it is 1 zero, then it is10 =1kloc-0/and three zeros, then it is 1000 = 10 3.

Therefore, 1999 zeros is 10 (1999).

Sn=n(a 1+an)/2

a 1= 1,n=an= 10^( 1999)

sn=( 10^ 1999)*[ 1+( 10^ 1999)]/2

=[ 10*( 10^ 1998)]*[ 1+( 10^ 1999)]/2

=( 10/2)*( 10^ 1998)*[ 1+( 10^ 1999)]

=5*( 10^ 1998)*[ 1+( 10^ 1999)]

={5*[ 1+( 10^ 1999)]}*( 10^ 1998)

Approximately 1+( 10 1999)

1+102 =1+100 =10/,and the number of zeros between two1=/kloc.

1+105 =1+100000 =10001,and the number of zeros between two1= 5-/kloc-0.

Therefore, 1+(101999) =10000 ... 000 (1998 zeros)1.

Since 5 * [1+(101999)] doesn't need to be carried, just change1to 5.

Therefore, the original formula =[5000...000( 1998 zeros)1] * (1kloc-0/998).