If n is an integer such that n > 9, which of the following could be

ShreyKapil08 yes, the estimation & the solution is right.

Yes, the method is correct and one of the right ways to do it.

I think Ans is D .

If n is an integer such that n > 9, which of the following could be the remainder when (2+2^2+2^3+2^4+....+2^n) is divided
by 3 ?

it is an GP. with common ratio 2.
so Sum = a (r^n-1) / (r-1) = 2 (2^n - 1) / (2-1) = 2 ^ (n+1) - 2 .

from here on we can test value of n = 10 , 11 , 12 , 13 .14

for n = 10 .
2^11 - 2 / 3 = 4 (9-1)^3 - 2 / 3 = final term contributing to reminder from binomial theorem 4(......+(-1)^3) -2 = -4 -2 = -6 which is divisible by 3 . So reminder is zero when n=10.

for n = 11 .
2 ^ 12 - 2 / 3 = (9-1)^4 -2 /3 = final term contributing to reminder from binomial theorem (.... + (-1)^4 -2) = 1-2 = -1 , which will produce reminder of -1 +3 = 2 .

similarly n=12 . reminder 0
n = 13.. rem 2
n = 14 .. rem 0 ..

so only possible reminders are 0 and 2 depending on the value of n .
So ans is D.

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