What will be the digit at the unit's place of $$1^3+2^3+3^3+4^3+5^3+6^3+7^3+8^3+9^3?$$​​

Correct option is C

Given:
Expression = $$1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3$$​​
Formula Used:
Sum of cubes of first n natural numbers = $$(\frac{n(n+1)}{2})^2$$​​
Solution:
The number of terms n is 9.
Sum = $$(\frac{9(9+1)}{2})^2$$​​

$$= (\frac{9 × 10}{2})^2$$​ $$= (45)^2$$​​
The unit digit of 45 is 5.
Since any positive integer power of a number ending in 5 also ends in 5, the unit digit is 5.
Final Answer
So the correct answer is (c)