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Question: Find the remainder when \[{1^{2013}} + {2^{2013}} + {3^{2013}}........ + {2012^{2013}}\]is divided b...

Find the remainder when 12013+22013+32013........+20122013{1^{2013}} + {2^{2013}} + {3^{2013}}........ + {2012^{2013}}is divided by 2013

Explanation

Solution

To find the remainder of the given series we treat the given series is like 1n+2n+...+(n−2)n+(n−1)n{1^n} + {2^n} + ... + {\left( {n - 2} \right)^n} + {\left( {n - 1} \right)^n}. Write the last term in the expanded form such that we will see there are some terms which will get canceled out and only multiple of n terms will be there. So we can easily find the value of remainder.

Complete step-by-step solution:
Considering the given expression,
1n+2n+...+(n−2)n+(n−1)n{1^n} + {2^n} + ... + {\left( {n - 2} \right)^n} + {\left( {n - 1} \right)^n} ………(i)
Starting at right hand side, expand the bracket,
(n−1)n=nn+...+(−1)n  {\left( {n - 1} \right)^n} = {n^n} + ... + {\left( { - 1} \right)^n}\;[ This is same as expanding (x+1)2  =x2+2x+12    {\left( {x + 1} \right)^{2\;}} = {x^2} + 2x + {1^{2\;}}\;we can observe that power of x is decreasing and at the end constant term is obtained]
Now each term (excluding last term) has a n in it (meaning no remainder as when it will be divided by n remainder will be 0) but remember we are excluding the last term.
Since here n = 2013, which is odd,
(−1)n  =−(1n)  {\left( { - 1} \right)^n}\; = - ({1^n})\;
i.e, (n−1)n  {\left( {n - 1} \right)^n}\;= (some multiple of n) - (1n)  ({1^n})\;
Similarly, the bracket term from R.H.S. ,
(n−2)n   = nn  + ... +(−2)n  {\left( {n - 2} \right)^{n\;}}{\text{ = }}{n^n}\; + {\text{ }}...{\text{ }} + {\left( { - 2} \right)^n}\;
(n−2)n  {\left( {n - 2} \right)^n}\;= terms of n + ... −(2n)   - ({2^n})\;
i.e, (n - 2)n = (some multiple of n) −(2n)   - ({2^n})\;
Now compare with the equation (i).
Clearly, the original numbers on the LHS are cancelling with the terms on the RHS because 1n in LHS will be cancelled with 1n in RHS and the same for 2 n and others too.
ie
{(1n)  ({1^n})\; + (some multiple of n) - (1n)  ({1^n})\;} +
{(2n)  ({2^n})\; + (some multiple of n) - −(2n)   - ({2^n})\;} +
....and so on
Since 2012 is even, these terms match exactly, giving a remainder of 0.

Note: Always make sure to use proper expansion for the given expression. If there are even numbers of terms present on this series then remainder would not be zero.