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Using Binomial Theorem, Prove that 2 3 N − 7 N − 1 is Divisible by 49, Where N ∈ N .

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Question

Using binomial theorem, prove that \[2^{3n} - 7n - 1\] is divisible by 49, where \[n \in N\] .

 
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Solution

\[2^{3n} - 7n - 1 = 8^n - 7n - 1\]       ...(1) 

\[Now, \]

\[ 8^n = (1 + 7 )^n \]

\[ =^{n}{}{C}_0 + ^{n}{}{C}_1 \times 7^1 + ^{n}{}{C}_2 \times 7^2 + ^{n}{}{C}_3 \times 7^3 + ^{n}{}{C}_4 \times 7^4 + . . . + ^{n}{}{C}_n \times 7^n \]

\[ \Rightarrow 8^n = 1 + 7n + 49[ ^{n}{}{C}_2 +^{n}{}{C}_3 \times 7^1 +^{n}{}{C}_4 \times 7^2 + . . . + ^{n}{}{C}_n \times 7^{n - 2} ]\]

\[ \Rightarrow 8^n - 1 - 7n = 49 \times \left( \text{ An integer}  \right)\]

\[\text{ Now, }  \]

\[ 8^n - 1 - 7n \text{ b is divisible by }49\]

\[\text{ Or, } \]

\[ 2^{3n} - 1 - 7 \text{ n is divisible by 49 } \left[ \text{ From } (1) \right]\]

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Introduction of Binomial Theorem
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Chapter 18: Binomial Theorem - Exercise 18.1 [Page 12]

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RD Sharma Mathematics [English] Class 11
Chapter 18 Binomial Theorem
Exercise 18.1 | Q 6 | Page 12

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