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What will be the digit at the unit's place of 13+23+33+43+53+63+73+83+93?1^3+2^3+3^3+4^3+5^3+6^3+7^3+8^3+9^3?13+23+33+43+53+63+73+83+93?​​
Question

What will be the digit at the unit's place of 13+23+33+43+53+63+73+83+93?1^3+2^3+3^3+4^3+5^3+6^3+7^3+8^3+9^3?​​

A.

7

B.

0

C.

5

D.

9

Correct option is C

Given:
Expression = 13+23+33+43+53+63+73+83+931^3 + 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 = (n(n+1)2)2(\frac{n(n+1)}{2})^2​​
Solution:
The number of terms n is 9.
Sum = (9(9+1)2)2 (\frac{9(9+1)}{2})^2​​

=(9×102)2 = (\frac{9 × 10}{2})^2​ =(45)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)

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