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The remainder, if 1+2+2^2+2^3+....+2^199...

The remainder, if `1+2+2^2+2^3+....+2^1999` in 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 PUBLICATION-BINOMIAL THEOREM-Numerical
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  3. If R is remainder when 6^(83)+8^(83) is divided by 49, then the value ...

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  5. Given (1-2x+5x^2-10x^3)(1+x)^n=1+a1x+a2x^2+.... and that a1^2=2a2, the...

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  11. The sum of the series (.^(101)C(1))/(.^(101)C(0)) + (2..^(101)C(2))/...

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  18. The value of sum(0leiltjle5) sum(""^(5)C(j))(""^(j)C(i)) is equal to "...

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  19. If (1-x-x^(2))^(20) = sum(r=0)^(40)a(r).x^(r ), then value of a(1) + 3...

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