We have:

9^2k = ...1

9^2k+1 = ...9

=> A = 1 + 9 + 9² + .... + 9⁵⁰

= 1 + 9 + ... 1 + .... + ...1

We easy see A will have sum A have 51 terms.

and 1 is first and last, so sum A have 26 last-digit terms 1 and 25 last-digit terms 9

So last-digit is: 26*1 + 25*9 = 251 = ...1

So last digits of 1 + 9 + ... + 9⁵⁰ is 1

How about this ?

\(A=1+9+9^2+...+9^{50}\)

\(\Rightarrow9A=9+9^2+9^3+...+9^{50}+9^{51}\)

\(\rightarrow9A-A=8A=\left(9+9^2+9^3+...+9^{50}+9^{51}\right)-\left(1+9+9^2+...+9^{50}\right)\)

\(\Rightarrow8A=9^{51}-1\Rightarrow A=\dfrac{9^{51}-1}{8}\)

We find that \(9^{2k},9^{2k+1},...\)always have the last digit is 1 or 9, or we can simply understand that whether 9 has index number is odd => the last digit is 9, and whether 9 has index number is even => the last number is 1.

\(9^{51}\) has the last digit is 9 (Cause 51 is odd number)

So : \(A=\dfrac{\overline{B9}-1}{8}=\dfrac{\overline{B8}}{8}=\overline{C1}\)

Therefore, the last digit of \(1+9+...+9^{50}\) is 1.

Answer : 1.