Question

The last digit of (1³+ 2³+ 3³+ 4³+ 5³+ 6³+ 7³+ 8³ + 9³ + 10³)¹⁰ is:


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Answer 1

Answer:

Formula for 1^3 + 2^3 + 3^3 + ……….. + n^3 is given as:

[n(n+1)/2]^2

So, 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3

= [7*(7+1)/2]^2

= [7*8/2]^2

= 28^2

= 784

The last digit is 4.

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Answer 2

Answer:

when the bracket is solved..

i.e { 11+8+27+64+125+216+343+512+729+1000}

we get 3025

now ,

${3025}^{10}$

as last digit is 5 in 3025..we will always get 5 in the end.