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Which of the following is true for the number $2^{4 n}-2^{n}(7 n+1)$ , where n is any positive Integer?
Option: 1
The highest divisor of the number among the given options is 7.

Option: 2
The highest divisor of the number among the given options is 7².

Option: 3
The highest divisor of the number among the given options is 14².

Option: 4
The highest divisor of the number among the given options is 14.

Answers (1)

best_answer

$\begin{array}{l}{2^{4 n}-2^{n}(7 n+1)=(16)^{n}-2^{n}(7 n+1)} \\ {\quad=(2+14)^{n}-2^{n} \cdot 7 n-2^{n}} \\ {\quad=\left(2^{n}+^{n} C_{1} 2^{n-1} \cdot 14+^{n} C_{2} 2^{n-2} \cdot 14^{2}+\ldots+14^{n}\right)-2^{n} \cdot 7 n-2^{n}} \\ {\quad=14^{2}\left(^{n} C_{2} 2^{n-2}+^{n} C_{3} 2^{n-3} 14+\ldots+14^{n-2}\right)+\left(2^{n}+^{n} C_{1} \cdot 2^{n-1} \cdot 14-2^{n} \cdot 7 n-2^{n}\right)} \\ {\quad=14^{2}\left(^n C_{2} 2^{n-2}+^{n} C_{3} 2^{n-3} \cdot 14+\ldots+14^{n-2}\right)+\left(2^{n}+n 2^{n-1} \cdot 2^{1} \cdot 7-2^{n} \cdot 7 n-2^{n}\right)} \\ {\quad=14^{2}\left(^{n} C_{2} \cdot 2^{n-2}+^{n} C_{3} \cdot 2^{n-3} \cdot 14+\ldots+14^{n-2}\right)}\end{array}$

This is divisible by 14²

Posted by

HARSH KANKARIA

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