Let Sn = an1+an2+…+ann be sum of numbers in nth row. Then, an1=ann=n-1 anr=an-1r-1+an-1r for r=2,…,n-1 So, Sn = 2(n-1) + sum{anr}r=2 to n-1 = 2(n-1) + sum{an-1r-1+an-1r}r=2 to n-1 = 2(n-1) + 2 sum{an-1r}r=1 to n-1 – an-1 1 – an-1 n-1 = 2(n-1) + 2Sn-1 – 2(n-2) = 2+2Sn-1 = 2+2(2+2Sn-2) = 2+2^2 + 2^2 Sn-2 = 2+2^2 + 2^2 (2+2Sn-3) = 2+2^2 + 2^3 +2^3Sn-3 … = 2+2^2+2^3+…+2^n-1+2^n-1S1 = 2+2^2+2^3+…+2^n-1 = 2(1-2^n-1)/(1-2) = 2[2^(n-1) – 1] S100 = 2[2^99-1] Let fn be last 2 digits of 2^n Since fn = f(20+n), so, f99=f19=88 Since 2(88-1)=174 Last 2 digits of s100=74