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Re: Unit digit of a series. [#permalink]
31 Jul 2010, 14:04
\(S(n)= 4^n +5^(n+1) +3\)
The way to solve this would be to look at each individual term. \(4^1 = 4\) \(4^2 = 16\) \(4^3 = 64\) and so on. So, the cyclicity of this is 2. Every 2 terms the last digit is 4 when the n is odd and 6 when n is even.
With 5, its always 5.
If n = 100, it's an even number and hence 4^100 will end in 6 and 5^101 will end in 5.
Hence the sum of last digits will be \(6+5+3 = 14\) so it'd be 4.