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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.

A

0

B

1

C

2

D

3

Text Solution

Verified by Experts

The correct Answer is:
a
We have , ` S= (1(2^(2000) -1))/(2-1) = 2^(2000) - 1 = (2^(2))^(1000) - 1`
` = (5-1)^(1000) -1`
` = (5^(1000- 1000)C_(1) *5^(999) + ""^(1000)C_(2) *5^(998) ...+ ""^(1000)C_(998) * 5^(2) - ""^(1000)C_(999) *5 + 1) - 1`
` =5 (5^(999) - ""^(1000)C_(1) *5^(998)+ ""^(1000)C_(2) * 5^(997) -...- ""^(1000)C_(999) )`
` therefore ` Remainder is 0.
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