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