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The remainder, if 1+2+2^2++2^(1999) is d...

The remainder, if `1+2+2^2++2^(1999)` is divided by 5 is.

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The correct Answer is:
0
`1+2+2^(2)+2^(3)+"……"+2^(1999)`
`= (1(2^(2000)-1))/(2-1)`
`= 2^(2000) - 1`
`= (1-5)^(1000) - 1`
`= 1- .^(1000)C_(1).5+.^(1000)C_(2).5^(2)+"……"+.^(1000)C_(1000).5^(1000)-1`
which is divisible by 5.
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CENGAGE ENGLISH-BINOMIAL THEOREM-Numerical
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