First slide

Binomial theorem for positive integral Index

Question

$The remainder when 2^{0} + 2^{1} + 2^{2} + 2^{3} + \dots \dots + 2^{2014} is divided by 9 is$

Moderate

A

7
B

8
C

4
D

3

Solution

$\begin{array}{l} Consider, 2^{0} + 2^{1} + 2^{2} + 2^{3} + \dots \dots . + 2^{2014} \\ = \frac{1 (2^{2015} - 1)}{2 - 1} = 2^{2015} - 1 (∵ In G.P series S_{n} = \frac{a (r^{n} - 1)}{r - 1}) \\ = 2^{2} {(2^{3})}^{671} - 1 \\ = 4 (9 - 1)^{671} - 1 \\ = 4 (9 λ - 1) - 1, λ \in ℝ \\ = 4 (9 λ) - 5 \\ = 4 (9 λ) - 9 + 4 \\ = 9 (4 λ - 1) + 4 \end{array}$
Since it is divided by 9, then reminder is 4.
Therefore, the correct answer is (3).

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