Simply Add

Sum of First 100 numbers = (0+1+2+...+99) = 1+2+3+...+99= 99×100/2 = 4950

or you know sum of 1 to 100 is 5050 then sum of 1 to 99 = 5050 – 100 = 4950

AP-----> 0,1,2,3,4...... since 1st term is 0 we can consider AP's 1st term as 1 therefore the sum of 1st 100 terms of original AP will be equal to sum of 1st 99 terms of considered AP ..
a=1(1st term) d=1(common difference) n=99(no.of terms)
S=sum
S(n)=n/2[2a+(n-1)d]
S(99)= 99/2[2(1)+(99-1)1]
=99/2[2+98]
=99/2[100]
=99*50
=4950

Summing all single digit units we get 45.

This repeats for 10s, 20s, 30s,...90s, for a total of 10 times (0s too).

Summing all 10s multiples we get 10+20+30...+90=450.

This repeats for 10 times for each 10s multiple (10,11,12...19; 20,21,22,...29; ...)

450*10=4500.

4500+450= $\boxed{4950}$ .