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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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