Find the last digit of this number:
$({}_{4n+1} C_0 )^{4m+1} + ({}_{4n+1} C_1 )^{4m+1} +({}_{4n+1} C_2 )^{4m+1} + \cdots + ({}_{4n+1} C_{4n+1} )^{4m+1}\;,$
where $n$, $m$ belong to the holy set of natural numbers.
Find the last digit of this number:
$({}_{4n+1} C_0 )^{4m+1} + ({}_{4n+1} C_1 )^{4m+1} +({}_{4n+1} C_2 )^{4m+1} + \cdots + ({}_{4n+1} C_{4n+1} )^{4m+1}\;,$
where $n$, $m$ belong to the holy set of natural numbers.
Here is my approach,
It is a fact that for any natural number n, n^4k+1 , has the same unit digit as n, itself, (for any natural number k)
so
in the series we just vanish all the powers of each terms, as we only have to find the unit digit,
and doing so gives us just the series representing sum of binomial coefficients of the series (1+x)^4n+1
whose sum is 2^4n+1
and last digit of 2^4n+1 is 2 itself, (declared earlier)
The ingredients you need to solve this are Euler's theorem (along with the value of Euler's totient function for the base of our decimal system) and the binomial theorem (applied to a power of $1+1$), or alternatively the fact that the total number of subsets of a $k$-element set is $2^k$.
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